This note is my bird’s eye view of analysis, espeically developing key ideas in metric spaces and continuity to make our life easier later. Analysis is about problems, counter-examples, and crafting of clever inequality among many other things (not that I would know :)) and we will do plenty of that to come

Analysis-0 - Metric Spaces

In order to do analysis, one needs a notion of distance, and one of the biggest traps is to blindly borrow intuition from $\mathbb{R}$ or $\mathbb{R}^2$, so we will borrow with our eyes open.

0 - Metric Spaces and Open Balls

a set $X$ and a map $d: X^2 \to \mathbb{R}_{\geq 0}$ is called a metric space if:

• $d(x,y) = 0 \iff x = y$ for any $x,y \in X$
• $d(x,y) = d(y,x)$ for any $x,y \in X$
• $d(x,z)+d(z,y) \ge d(x,y)$ for any $x,y,z \in X$ (triangle inequality)

We will work with two objects very often, one is open balls, and the other is punctured open balls:

let $X$ be a metric space, for any $x_{0} \in X$ and any $r > 0$, the set $B_{r}(x_{0}) := \{x\in X: d(x,x_{0}) < r\}$ is called an open ball (or open neighbourhood) of $x_{0}$ with radius $r$.

The set $B_{r}^\times(x_{0}):= B_{r}(x_{0}) - \{x_{0}\}$ we will call a punctured open ball. Notice that an open ball is always nonempty, (it has its centre at least) whereas the punctured open ball may be empty.

Indeed, if we have a “set without its boundary” intuitively, we know that any point inside, will have some (perhaps really small) open ball around it, completely contained inside that set.

Let $U \subset X$, we say $u \in U$ is an interior point if there exists some $\epsilon > 0$ for which $B_{\epsilon}(u) \subset U$.

We say $U$ is open if every $u \in U$ is an interior point of $U$.

if $X$ is a metric space, we see (vacuously) that $\emptyset$ is open, and $X$ is open too.

Extremal Examples

Now, in analysis, we want to hunt for extremal examples. Is there a metric space where all subsets are open? consider $\mathbb{Z}$ with the usual metric $d(x,y) = |y-x|$. Choose $\epsilon = 0.01$, now let $U \subset \mathbb{Z}$ and $u \in U$, notice that $B_{\epsilon}(u) = \{u\} \subset U$.

Is there a metric space where the only open sets are the trivial ones? suppose $X$ had at least two points $x,y$, and let $d^* = d(x,y) > 0$. then $B_{d^*}(x)$ is nonempty since it contains $x$, and it is not all of $X$ since it does not contain $y$. thus $X$ has a nontrivial open set.

therefore, the metric topology on $X$ is trivial if and only if $|X| \leq 1$.

1 - Limit Points and Closed Sets

The next object of concern is limit points. let $X$ be a metric space and $U \subset X$. we say $p \in X$ is a limit point, if for any $\epsilon > 0$, $B^\times_{\epsilon}(p) \cap U \neq \emptyset$. That is, any punctured ball around $p$ has elements in $U$.

It is quite evident that if $p$ is a limit point of $U$ any ball around $p$ has an infinite number of elements in $U$.

For the sake of contradiction, suppose for some $\epsilon^* > 0$, $B^\times_{\epsilon^*}(p) \cap U$ has only a finite number of elements: $u_{1},u_{2},\dots,u_{n}$. set $\gamma = \min(d(p,u_{1}),d(p,u_{2}),\dots,d(p,u_{n}))$.

Now notice that $B^\times_{\gamma}(p) \cap U \subset B^\times_{\epsilon^*}(p) \cap U$, but by the choice of $\gamma$, none of $u_1,\dots,u_n$ is in $B^\times_{\gamma}(p)$. thus, $B^\times_{\gamma}(p) \cap U = \emptyset$, which contradicts the fact that $p$ is a limit point of $U$.

we say a set is closed, if it contains all its limit points.

Closed Sets and Open Complements

Now, we shall prove here, as a theorem, what is taken as a definition in topology.

let $X$ be a metric space and $U \subset X$, then $U$ is closed if and only if $X-U$ is open.

suppose $U$ is closed, and let $x \in X-U$. since $U$ contains all its limit points, $x$ is not a limit point of $U$. thus, for some $r > 0$, $B^\times_r(x) \cap U = \emptyset$. since $x \notin U$, we also have $B_r(x) \cap U = \emptyset$, ie $B_r(x) \subset X-U$. that is $x$ is interior in $X-U$. thus $X-U$ is open.

The other direction is very straightforward. Suppose $X-U$ is open, and $p$ is a limit point of $U$, further suppose $p \not \in U$, thus $p$ is interior to $X-U$, contradicting that $p$ is a limit point of $U$.

2 - The Metric Topology and Continuous Maps

The next claim we will make is that open balls, are the basis which generate the metric topology of a metric space.

What do we require of basis? we need them to be open and cover the space, we need for any point in the intersection of two basis another basis contained in that intersection containing the point. Such things are satisfied by open balls.

Next we will show that any open set is a union of open balls. This is true because in the metric space definition, every point inside an open set is interior, thus has an open ball completely contained in the open set. Thus we can take the union of all such balls.

Open Sets Form a Topology

Next we will show that open sets in the metric sense form a topology: let $X$ be a metric space: of course $\emptyset, X$ are open.

now, suppose $U_{\alpha}$, $\alpha \in A$ is a collection of open sets. let $u \in \bigcup_{\alpha \in A} U_{\alpha}$, of course, for some $\alpha^*$, $u \in U_{\alpha^*}$, ie some $B_{r}(u) \subset U_{\alpha^*}$. Therefore, obviously, $B_{r}(u) \subset \bigcup_{\alpha \in A} U_{\alpha}$, thus $u$ is interior in the union of that collection.

now let $U_{i}$, $1 \leq i \leq n$ be a finite collection of open sets. let $u \in \bigcap_{i=1}^nU_{i}$, of course then, there exists $r_{i}$, $1 \leq i \leq n$ such that $B_{r_{i}}(u) \subset U_{i}$.

now, let $r^* = \min(r_{1},r_{2},\dots,r_{n})$, then for any $i$, $B_{r^*}(u) \subset B_{r_i}(u) \subset U_{i}$.

ie, $B_{r^*}(u) \subset \bigcap_{i=1}^nU_{i}$.

Continuous Maps Between Metric Spaces

Now, if $f: X \to Y$ is a map between metric spaces, we have two definitions of continuity, and they better be equivalent:

• $f$ is continuous if pre-images of open sets are open, (in the metric topology).
• for each $x_{0} \in X$ and each $\epsilon > 0$, there exists $\delta > 0$ such that $d_{X}(x,x_{0})< \delta$ implies $d_{Y}(f(x), f(x_{0})) < \epsilon$.

suppose $f$ is continuous as given by the first definition.

let $x_{0} \in X$, consider any $B_{\epsilon}(f(x_{0}))$. Indeed, $f^{-1}(B_{\epsilon}(f(x_{0})))$ is open in $X$, ie $x_{0}$ is interior to it. that is, there exists some $\delta > 0$ such that $B_{\delta}(x_{0}) \subset f^{-1}(B_{\epsilon}(f(x_{0})))$, ie $f(B_{\delta}(x_{0})) \subset B_{\epsilon}(f(x_{0}))$.

Now suppose $f$ is continuous as given by the second definition. let $U$ be open in $Y$. let $v \in f^{-1}(U)$, now we know that for some $\epsilon > 0$, $B_{\epsilon}(f(v)) \subset U$. for that $\epsilon$, there is some $\delta > 0$ such that $f(B_{\delta}(v)) \subset B_{\epsilon}(f(v)) \subset U$. of course then $B_{\delta}(v) \subset f^{-1}(U)$, thus $f^{-1}(U)$ is open.

Now, it is quite clear, using complements, that in topological spaces (and thus metric spaces) arbitrary intersections and finite unions of closed sets are closed.

3 - Dense, Perfect, and Compact Sets

Now some more definitions:

A subset $U$ of a metric space $X$ is said to be dense, if for each $x \in X$, either $x \in U$ or $x$ is a limit point of $U$, moreover if $X$ is a metric space such that any $x \in X$ is a limit point of $X$, then the space is said to be dense in itself.

Now, if $u \in U$ is not a limit point of $U$, then it is called an isolated point of $U$.

Finally $U$ is perfect if it is closed and every $u \in U$ is a limit point of $U$. ie $U$ is perfect if it is closed and has no isolated points.

As with compactness in topological spaces, a subset $U \subset X$ is compact, if every open cover of $U$ has a finite subcover.

note: Most of the properties on a subset of a space, (topological or metric) can be pulled directly down into the subspace induced by that subset, either by taking intersection of open sets, or by inheriting the metric. For example, a space $X$ is compact if every open cover has a finite subcover, $U$ is a compact subset if and only if as a subspace, it is a compact space.

We say that the empty set is bounded. A nonempty subset $U$ of a metric space $X$ is bounded if there exists $M > 0$ and some $u_{0} \in U$ such that for any $u \in U$, $d(u, u_{0}) \le M$.

Compact Sets Are Closed and Bounded

Suppose $X$ is a metric space and $U$ is a compact subset of $X$, then $U$ is closed and bounded:

first we will show that $U$ bounded. this is immediate if $U$ is empty, so suppose it is nonempty and take some $u_{0} \in U$. now build a sequence of balls $B_{r_{i}}(u_{0})$, $i \in \mathbb{N}$, $r_{i} = 2^i$. since the distance between any two points is finite, this sequence of balls covers $U$. Of course, it has a finite subcover, but every ball is contained in the next one in this sequence, thus there exists some $n \in \mathbb{N}$ such that $U \subset B_{2^n}(u_{0})$.

Next we will show that $U$ is closed, ie $X-U$ is open: this is immediate if $U$ is empty, so suppose it is nonempty and let $x \in X-U$. Now, build a cover of $U$ by taking $B_{r(u)}(u)$, where $r(u) = d(u,x)/3$. This choice ensures that even if the whole $r(u)$ ball sticks out toward $x$, every point in it remains more than $2d(u,x)/3$ away from $x$.

We will now prove this. select $r(u) = d(u,x)/3$ and get the finite subcover $B_{r(u_{i})}(u_{i})$, $1 \le i \le n$. now, let $r^* = \min(r(u_{i}), i \in [1,n])$ we claim that no $u \in U$ is in $B_{r^*}(x)$. let $u \in U$, then for some $i$, $u \in B_{r({u_{i}})}(u_{i})$. thus, $d(u_{i}, x) \le d(u,x) + d(u,u_{i})$. ie, $d(u,x) \ge d(u_{i},x) - d(u,u_{i})$. since $d(u,u_{i}) < r(u_{i})$ and $d(u_{i},x) = 3r(u_{i})$ we have $d(u,x) > 2r(u_{i}) \ge 2r^* > r^*$.

Continuous Images and Closed Subsets

The following holds for any topological space: let $f: X \to Y$ be continuous, and $X$ compact. then $f(X)$ is a compact subspace of $Y$.

Take an open cover of $f(X)$, take its pre-image under $f$ to get an open cover of $X$, since $X$ is compact, you get its finite subcover, send it back by taking its image under $f$, to get a finite subcover of $f(X)$.

lemma: let $X$ be a topological space, and $K \subset X$ a compact set. Then, if $U \subset K$ is closed, then $U$ is compact.

we know $K-U$ is open in the subspace $K$. Take any open cover of $U$, and extend it to an open cover of $K$ by adding $K-U$. Take a finite subcover of $K$, and discard $K-U$ if it was selected. what remains is a finite subcover of $U$.

Notice that any point of $\mathbb{Z}$ with the usual metric is isolated in $\mathbb{Z}$, and any point of $\mathbb Q$ is a limit point of itself. Ie $\mathbb Q$ is dense in itself.

4 - Ordered Fields and the Real Numbers

A field $F$ with an order $\le$ on the set $F$ is said to be an ordered field if the field operations play nice with order:

• $x \le y$, $x' \le y'$ implies $x + x' \le y + y'$
• $0 \le x$, $0 \le y$ implies $0 \le xy$

ordered fields are cute, for example, as a field we know $1 \neq 0$ suppose $1 \le 0$, and of course $-1 \le -1$ (order is reflexive) thus we obtain $0 \le -1$ by adding them together. now we can multiply two copies of that inequality, and get $0 \le (-1)(-1) = 1$. together with $1 \le 0$, antisymmetry would give $1=0$, a contradiction. (these things follow from field axioms)

that is $0 \le 1$ in an ordered field.

an ordered set $S$ has the least upper bound property, or the supremum property, if any nonempty subset with an upper bound, has the least upper bound defined as follows:

let $U \subset S$ we say that $U$ is bounded above if for some $a \in S$, $u \le a$ for any $u \in U$. Further, we call $a^* \in S$ the least upper bound of $U$ if $a^*$ is an upper bound of $U$, and for any upper bound $a$ of $U$, we have $a^* \le a$.

We can analogously define lower bounds and greatest lower bounds, and the greatest lower bound property on an ordered set.

We define $\mathbb{R}$ to be the ordered field which has the supremum property and contains $\mathbb Q$ as a subfield (up to isomorphism), and we will not bother constructing it. Further, it follows that $\mathbb{R}$ has the infimum property as well, that is, any nonempty set with a lower bound, has the greatest lower bound.

If nothing is mentioned, we take the usual $d(x,y) = |y-x|$ metric on $\mathbb{R}$.

lemma -1: let $U \subset \mathbb{R}$ be nonempty and bounded above and $\alpha = \sup(U)$. Then $\alpha \in U$ or $\alpha$ is a limit point of $U$ (perhaps both). proof: suppose neither is true. well since $\alpha$ is not a limit point of $U$, there exists some $\epsilon > 0$ such that $(\alpha -\epsilon, \alpha +\epsilon) \cap U$ is empty. ie, for each $u \in U$, $u \le \alpha-\epsilon$ (of course $u \le \alpha$). thus, $\alpha - \epsilon$ is an upper bound of $U$, a contradiction.

lemma 0: $[0,1]$ is closed.

proof: we will show $\mathbb{R}-[0,1] = (-\infty,0) \cup (1, \infty)$ is open. of course any point in $\mathbb{R} -[0,1]$ is interior to it.

lemma 1: $[0,1]$ is bounded. (left as an exercise for the reader)

Compactness of the Unit Interval

Theorem. $[0,1]$ is a compact subspace of $\mathbb{R}$.

proof: This one is a banger. Let $\mathscr{U}$ be an open cover of $[0,1]$ clearly, some $U \in \mathscr{U}$ contains $0$. Of course then $0$ is interior to $U$, ie for some $\epsilon > 0$, $[0,\epsilon) \subset U$. so at least $[0,\epsilon/2]$ has a finite subcover. we would like to extend him :) (for example, $\epsilon/2$ in some $U \in \mathscr{U}$ and thus he is interior to $U$, and …)

let $S = \{ x \in [0,1]: [0,x] \subset \bigcup_{i=1}^n U_{i}, \ U_{i} \in \mathscr{U} \}$. That is $S$ is the subset of $[0,1]$ whose closed interval starting at $0$ is finitely covered by sets in $\mathscr{U}$. $S$ is nonempty and bounded above, thus let $\alpha =\sup(S)$. We claim $\alpha = 1$. For contradiction, suppose $\alpha < 1$. that is, $\alpha$ is in some $U_{\alpha} \in \mathscr{U}$ and is interior to it. thus, for some $\epsilon > 0$ with $\alpha+\epsilon < 1$, $(\alpha-\epsilon, \alpha + \epsilon) \cap [0,1] \subset U_{\alpha}$.

Now, if $\alpha \in S$, then of course $[0,\alpha]$ is finitely covered, and hence $[0,\alpha+ \epsilon)$ is finitely covered by taking union with $U_{\alpha}$. of course then $[0, \alpha + (\epsilon/2)]$ is finitely covered, ie $\alpha + (\epsilon/2) \in S$, a contradiction.

Otherwise, $\alpha$ is a limit point of $S$. thus, there is some $s \in S$ with $\alpha-(\epsilon/2) < s < \alpha$. Ie, $[0,s]$ is finitely covered, and taking union with $U_{\alpha}$, $[0, \alpha +(\epsilon/2)]$ is yet again finitely covered, a contradiction.

Thus, $\alpha = 1$. well cool, we still don’t know if $1 \in S$, (if $1 \in S$, we are done so suppose otherwise) but what we do know is that since $1 \in U$ for some guy in the open cover, for some $\delta > 0$, we have $(1-\delta,1] \subset U$. since $1$ is a limit point of $S$, there is some $s \in S$ with $1-\delta < s < 1$. ie $[0,s]$ is finitely covered, and taking union with $U$, $[0,1]$ is finitely covered.

5 - The Standard Topology on $\mathbb{R}^n$

Now, we will pause and gain some intuition on the standard topology (using the $L_{2}$ or whatever the usual metric) on $\mathbb{R}^n$.

The projection of an open ball in $\mathbb{R}^n$ onto each axis gives an open interval. Let us show this:

let $\pi_{i}: \mathbb{R}^n \to \mathbb{R}$ be the projection map onto the $i$-th axis. that is, $\pi_{i}(\mathbf{x}) = \mathbf{x} \cdot \mathbf{e^i}$ where $\mathbf{e^i}$ is the usual $i$th basis.

let $\mathbf{x_{0}} \in \mathbb{R}^n$, and $r > 0$, we claim that $\pi_{i}(B_{r}(\mathbf{x_{0}})) = B_{r}(\mathbf{x_{0}^i})$.

First, let $L_{i} = \{ \mathbf{x} \in \mathbb R^n : \mathbf{x} = \mathbf{x_{0}} + t\mathbf{e^i}, t \in (-r,r) \}$ of course $L_{i} \subset B_{r}(\mathbf{x_{0}})$. Notice if $\mathbf{x} \in L_{i}$, $\pi_{i}(\mathbf{x}) = \mathbf{x_{0}}\cdot \mathbf{e^i} + t = \mathbf{x_{0}^i} + t$ where $t \in (-r,r)$ thus, $\pi_{i}(L_{i}) = B_{r}(\mathbf{x_{0}^i})$.

This means that $B_{r}(\mathbf{x_{0}^i}) \subset \pi_{i}(B_{r}(\mathbf{x_{0}}))$. But of course, if $\mathbf{x} \in B_{r}(\mathbf{x_{0}})$, then $|\mathbf{x^i}-\mathbf{x_{0}^i}| \le |\mathbf{x}-\mathbf{x_{0}}| < r$. That is, $\pi_{i}(\mathbf{x}) = \mathbf{x^i} \in B_{r}(\mathbf{x_{0}^i})$.

thus, we have $\pi_{i}(B_{r}(\mathbf{x_{0}})) \subset B_{r}(\mathbf{x_{0}^i})$.

Now, we claim that $B_{r}(\mathbf{x_{0}}) \subset \prod_{i=1}^n \pi_{i}(B_{r}(\mathbf{x_{0}}))$. That is, the intervals obtained by projecting an open ball onto each axis, when taken product contain the open ball.

That is, $B_{r}(\mathbf{x_{0}}) \subset \prod_{i=1}^n B_{r}(\mathbf{x_{0}^i})$.

As before, if $\mathbf{x} \in B_{r}(\mathbf{x_{0}})$ then $\mathbf{x} = \mathbf{x_{0}} + t\mathbf{u}$ where $\mathbf{u}$ is a unit vector and $t \in (-r,r)$.

but of course $\mathbf{x} = <\mathbf{x^i}>$ (as a tuple), where $\mathbf{x^i} = \mathbf{x} \cdot \mathbf{e^i} \in B_{r}(\mathbf{x_{0}^i})$ as shown earlier. Thus by definition, $\mathbf{x} \in \prod_{i=1}^n B_{r}(\mathbf{x_{0}^i})$.

And finally, we will show that there exists a SINGLE $\epsilon$ (this will help A LOT in showing continuity, I think) for which $\prod_{i=1}^n B_{\epsilon}(\mathbf{x_{0}^i}) \subset B_{r}(\mathbf{x_{0}})$.

Indeed, if $\mathbf{x} \in \prod_{i=1}^n B_{\epsilon}(\mathbf{x_{0}^i})$, then $\mathbf{x^i} \in (\mathbf{x_{0}^i}-\epsilon, \mathbf{x_{0}^i} + \epsilon)$, that is

$$-\epsilon < \mathbf{x^i} - \mathbf{x_{0}}^i < \epsilon.$$

equivalently,

$$\sum_{i=1}^n (\mathbf{x^i} - \mathbf{x_{0}^i})^2 < n\epsilon^2.$$

of course then, it is sufficient if $n\epsilon^2 < r^2$, i.e. $\epsilon < r/\sqrt{n}$.

Okay, very nice. Now, we shall show a couple of things.

let $U_{i}$, $1 \leq i \leq n$ be a collection of open sets in $\mathbb{R}$ we claim $\prod_{i=1}^n U_{i}$ is open in $\mathbb{R}^n$

let $\mathbf{x} \in \prod_{i=1}^n U_{i}$, then by definition, $\mathbf{x^i} \in U_{i}$, ie for some $\epsilon_{i} > 0$ $B_{\epsilon_{i}}(\mathbf{x^i}) \subset U_{i}$, let $\epsilon = \min_{i}(\epsilon_{i})$

then, $B_{\epsilon}(\mathbf{x^i}) \subset U_{i}$, thus $\prod_{i=1}^n B_{\epsilon}(\mathbf{x^i}) \subset \prod_{i=1}^n U_{i}$.

But as shown earlier, $B_{\epsilon}(\mathbf{x}) \subset \prod_{i=1}^n B_{\epsilon}(\mathbf{x^i})$ thus $\mathbf{x}$ is interior to $\prod_{i=1}^n U_{i}$.

Similarly, if $U \subset \mathbb{R}^n$ is open, each $U_{i} = \pi_{i}(U)$ is open in $\mathbb{R}$.

suppose $x \in \pi_{i}(U)$. then there is $\mathbf{x} \in U$ for which $\mathbf{x^i} = x$. Then for some $r > 0$, $B_{r}(\mathbf{x}) \subset U$ but of course then, $\pi_{i}(B_{r}(\mathbf{x})) \subset \pi_{i}(U)$. Equivalently, $B_{r}(\mathbf{x^i}) \subset U_{i}$.

In $\mathbb{R}^n$ we can thus denote the open box $\prod_{i=1}^n (\mathbf{x^i}-\epsilon, \mathbf{x^i} +\epsilon)$ as $I^n_{\epsilon}(\mathbf{x})$, and equivalently, we can say that $U \subset \mathbb{R}^n$ is open if and only if, for each $\mathbf{x} \in U$, there exists some $\epsilon > 0$ such that $I^n_{\epsilon}(\mathbf{x}) \subset U$.

Compact Boxes

Theorem. $[0,1]^n$ is a compact subspace of $\mathbb{R}^n$.

We will proceed by induction, (notice we have already shown the base case) Suppose $[0,1]^k$ is compact, $[0,1]^{k+1} = [0,1]^k \times [0,1]$ (up to isomorphism).

let $x \in [0,1]$ and consider the set $A_{x} = [0,1]^k \times \{x \}$. Notice that the set isomorphism $g: [0,1]^k \to [0,1]^k \times \{x\}$ given by $\mathbf{x} \mapsto (\mathbf{x},x)$ (concatenation) is continuous. Thus as shown earlier, $g([0,1]^k) = A_{x}$ is a compact subspace of $\mathbb{R}^{k+1}$.

Now let $\mathscr{V}$ be an open cover of $[0,1]^{k+1}$. of course, $\mathscr{V}$ is an open cover of $A_{x}$, and from that view, let $\mathbf{V}$ be a finite subcover of $A_{x}$.

since each $\mathbf{u} \in A_{x}$ is in some open set in $\mathbf{V}$, there exists $r(\mathbf{u})$ such that the open box $I^{k+1}_{r(\mathbf{u})}(\mathbf{u})$ is contained in some set in $\mathbf{V}$. but of course, the collection of all such open boxes cover $A_{x}$ and thus, there is a finite subcover $\mathscr{I} = \{ I^{k+1}_{r_{i}}(\mathbf{u_{i}}): 1 \le i \le t \}$, moreover $\mathbf{V}$ covers $\bigcup \mathscr{I}$.

The nice thing with open boxes is, it lets us reason about each axis independently.

Any $\mathbf{u} \in A_{x}$ is in some $I^{k+1}_{r_{i}}(\mathbf{u_{i}})$. of course, $\mathbf{u} = (\mathbf{u'},x)$ where $\mathbf{u'} = <\mathbf{u^i}>_{i=1}^k$.

thus, $\{\mathbf{u'}\}\times(x-r_{i},x+r_{i}) \subset I_{r_{i}}^{k+1}(\mathbf{u_{i}})$.

That is to say, taking $r^* = \min(r_{i})$, $[0,1]^k \times (x-r^*, x+ r^*)$ is covered by $\mathscr{I}$ and thus covered by $\mathbf{V}$. (note: the keen reader might yell at me saying hey $x = 0$ or $x = 1$ means the interval $(x-r,x+r)$ gets cutoff. if you like, you may take intersection with $[0,1]$, or work in the ambient space)

Indeed, then for any open cover $\mathscr{V}$ of $[0,1]^{k+1}$ and any $x \in [0,1]$, there exists some $r_x > 0$ for which a finite subcollection $\mathbf{V}_{x}$ of $\mathscr{V}$ covers $[0,1]^k \times (B_{r_x}(x) \cap [0,1])$.

but of course, the collection $\{B_{r_x}(x) \cap [0,1] : x \in [0,1]\}$ covers $[0,1]$, so it has a finite subcover $B_{r_{x_j}}(x_j) \cap [0,1]$, $1 \le j \le p$. The union of the corresponding finite collections $\mathbf{V}_{x_j}$ is then a finite subcover of $[0,1]^{k+1}$ by sets in $\mathscr{V}$.

Now we get a bunch of stuff for free. For example, $\prod_{i=1}^n [a_{i},b_{i}]$ is a compact subspace of $\mathbb{R}^n$.

owing to the lemma from before it is sufficient to show that the map $f: [0,1]^n \to \prod_{i=1}^n[a_{i},b_{i}]$ given by $\mathbf{u} \mapsto \mathbf{a} + (\mathbf{b}-\mathbf{a}) \mathbf{u}$ is continuous, (where $\mathbf{a}\cdot \mathbf{e^i} = a_{i}$ and $\mathbf{b} \cdot \mathbf{e^i} = b_{i}$, and the multiplication is elementwise) for any real $a_{i} \le b_{i}$.

If $\mathbf{b}=\mathbf{a}$, then $f$ is constant and therefore continuous. Otherwise, let $\epsilon > 0$, and let $|\mathbf{x} - \mathbf{x_{0}}| < \epsilon/|\mathbf{b}-\mathbf{a}|$. Notice $|f(\mathbf{x})- f(\mathbf{x_{0}})| = |(\mathbf{b}-\mathbf{a})(\mathbf{x} - \mathbf{x_{0}})| \le |\mathbf{b}-\mathbf{a}||\mathbf{x}-\mathbf{x_{0}}|$ (the right hand side is Hölder’s inequality, which we shall show below) and thus $|f(\mathbf{x})- f(\mathbf{x_{0}})| < \epsilon$.

Hölder’s Inequality

In $\mathbb{R}^n$, $|\mathbf{a} \mathbf{b}| \le |\mathbf{a}||\mathbf{b}|$ where the multiplication on vectors is elementwise. i.e. we must show

$$\sqrt{ \sum_{i=1}^n(a_{i}b_{i})^2 } \le \sqrt{ \sum_{i=1}^na_{i}^2 } \sqrt{ \sum_{i=1}^n b_{i}^2 }.$$

of course, we can square both sides, and show

$$\sum_{i=1}^n(a_{i}b_{i})^2 \le \sum_{i=1}^na_{i}^2 \sum_{i=1}^n b_{i}^2.$$

for ease, let $A_{n} = \sum_{i=1}^n a_{i}^2, \ B_{n} = \sum_{i=1}^n b_{i}^2$.

Notice

$$\begin{align*} A_{n+1}B_{n+1} &= A_{n+1}B_{n} + A_{n+1}b_{n+1}^2 \\ &= A_{n}B_{n} + a_{n+1}^2B_{n} + b_{n+1}^2A_{n} + a_{n+1}^2 b_{n+1}^2 \\ &\ge A_{n}B_{n} + a_{n+1}^2b_{n+1}^2. \end{align*}$$

Now proceed by induction, (the base case is obvious) and suppose $\sum_{i=1}^n(a_{i}b_{i})^2 \le A_{n}B_{n}$, of course $\sum_{i=1}^{n+1}(a_{i}b_{i})^2 \le A_{n}B_{n} + a_{n+1}^2b_{n+1}^2 \le A_{n+1}B_{n+1}$.

The Heine-Borel Theorem

Theorem. Every closed bounded subset in $\mathbb{R}^n$ is compact.

let $U \subset \mathbb{R}^n$ be closed and bounded. Of course then, $U \subset \prod_{i=1}^n[a_{i},b_{i}]$, and thus is a closed subset of a compact set.

Ie together, in $\mathbb{R}^n$, a subset is compact if and only if it is closed and bounded.

6 - Continuous Maps Between Real Vector Spaces

From our discussions earlier, it is quite clear that open boxes form an alternative basis, we further claim that if $f: \mathbb{R}^m \to \mathbb{R}^n$ has the property that for each $x \in \mathbb{R}^m$ and any $\epsilon > 0$, there exists $\delta > 0$ such that $f(I^m_{\delta}(x)) \subset I^n_{\epsilon}(f(x))$, then $f$ is continuous. This follows directly from seeing that if $U \subset \mathbb{R}^n$ is open, then for any $x \in f^{-1}(U)$, we can take an open $\epsilon$ box around $f(x)$ take its pre-image, which would contain the $\delta$ box around $x$. In fact, this is an equivalent definition of continuity of $f$.

we will use this to build a more workable, but equivalent condition to characterise continuous maps from any two real vector spaces.

Consider $f: \mathbb{R}^m \to \mathbb{R}^n$, we claim that $f$ is continuous if and only if each projection of $f$, $f_{i} = \pi_{i} \circ f: \mathbb{R}^m \to \mathbb{R}$ is continuous.

We only need to show one direction, as $\pi_{i}$ is continuous.

Now suppose each $f_{i}$ is continuous. fix $x \in \mathbb{R}^m$ and let $\epsilon > 0$. there exists some $\delta_{i}$ such that $f_{i}(I^m_{\delta_{i}}(x)) \subset (f_{i}(x) - \epsilon, f_{i}(x) + \epsilon)$.

choose $\delta = \min(\delta_{i})$ then $\pi_{i} \circ f(I^m_{\delta}(x)) \subset (f_{i}(x) - \epsilon, f_{i}(x) + \epsilon)$.

that is, $f(I^m_{\delta}(x)) \subset \pi_{i}^{-1}(f_{i}(x)-\epsilon, f_{i}(x) + \epsilon) = \mathbb{R}^{i-1} \times (f_{i}(x)-\epsilon, f_{i}(x) + \epsilon)\times \mathbb{R}^{n-i}$.

If a set is a subset of many sets, it is a subset of their intersection.

Thus, clearly, $f(I^m_{\delta}(x)) \subset \prod_{i=1}^n (f_{i}(x) -\epsilon, f_{i}(x)+ \epsilon) = I^n_{\epsilon}(f(x))$.

Now one might be tempted to say that $f: \mathbb{R}^m \to \mathbb{R}$ having $f_{j}: x_{j} \mapsto f(x_{1},x_{2},\dots x_{j},\dots, x_{m})$ continuous for each $j$ and each fixed choice of the rest, means that $f$ is continuous. This is not true, we will provide many counterexamples later :)

The Reals as a Topological Field

We have special topological spaces, with some algebraic structure on them, for which the operations are continuous. (check algebra note)

we claim $\mathbb{R}$ is a topological field.

we will show that the addition $+ : \mathbb{R}^2 \to \mathbb{R}$ is continuous. Let $\epsilon > 0$,

suppose $(x,y) \in I_{\epsilon/2}^2((x_{0},y_{0}))$ then $|x-x_{0}| < \epsilon/2$ and $|y-y_{0}| < \epsilon/2$. Now notice that $|(x+y) - (x_{0} + y_{0})| \le |x-x_{0}| + |y-y_{0}| < \epsilon$.

Next we will show that the multiplication $\cdot : \mathbb{R}^2 \to \mathbb{R}$ is continuous. let $\epsilon > 0$

now, suppose $|x-x_{0}|, |y-y_{0}| < \delta$. first notice $(x-x_{0})(y-y_{0}) = xy + x_{0}y_{0} -(x_{0}y + y_{0}x)$

that is, $(x-x_{0})(y-y_{0}) + x_{0}(y-y_{0}) + y_{0}(x-x_{0}) = xy -x_{0}y_{0}$.

that is, repeatedly using $|a+b| \le |a| + |b|$ and simplifying,

$$|xy-x_{0}y_{0}| < \delta^2 + \delta(|x_{0}| + |y_{0}|).$$

ie we demand $\delta < 1$ to make this chill. and sufficiently, we can have $\delta(|x_{0}|+|y_{0}| + 1) < \epsilon$

together, it is sufficient to take $\delta < \min(1, \epsilon/(|x_{0}| + |y_{0}| + 1))$.

Now alarm bells might be ringing over your head, don’t worry! of course, after taking an $\epsilon$ ball around $f(x_{0},y_{0})$ the $\delta$ ball around $(x_{0},y_{0})$ absolutely can depend on it!, of course it can’t depend on some point in the ball that isn’t the centre, because that doesn’t make sense, in order to specify elements of a ball, you must pick a radius, so that radius can’t depend on elements of the ball other than the centre. (which is common for all balls)

Finally, we will show that the inverse $^{{-1}}: \mathbb{R}^\times \to \mathbb{R}^\times$ is continuous.

first notice $|x_{0}|-|x| \le |x - x_{0}|$ (another variant of triangle inequality) thus $|x| \ge |x_{0}| -|x-x_{0}|$. now suppose $|x-x_{0}| < \delta < |x_0|$. then $|x| > |x_{0}| - \delta$, ie

$$|x x_{0}|^{-1} <1/(|x_{0}|(|x_{0}| - \delta)).$$

equivalently,

$$|x^{-1} - x_{0}^{-1}| < \delta/(|x_{0}|(|x_{0}| - \delta)).$$

sufficiently, we want $\delta < \epsilon|x_{0}|^2 -\epsilon |x_{0}|\delta$ thus $\delta < \epsilon|x_{0}|^2/(1+\epsilon|x_{0}|)$.

As shown in the algebra notes, thus the collection of continuous maps from any topological space to the reals form a ring under pointwise addition and multiplication. (the maps which do not take zero values are closed under pointwise multiplication and inverse, but not under addition, so they do not form a field.)

$\mathbb{R}^m$ as a Topological Vector Space

Next we shall show that $\mathbb{R}^m$ (over $\mathbb{R}$) forms a topological vector space, that is, addition and scalar products are continuous. (of course using the product topology wherever it’s ambiguous)

This follows directly, notice that $+ : \mathbb{R}^{2m} \to \mathbb{R}^m$ has each projection $\pi_{i} \circ + : (\mathbf{x},\mathbf{y}) \mapsto \mathbf{x^i} + \mathbf{y^i}$ continuous, (this is the addition in reals) similarly, the scalar multiplication $\cdot : \mathbb{R} \times \mathbb{R}^m \to \mathbb{R}^m$ has each projection $\pi_{i} \circ \cdot : (\lambda, \mathbf{x}) \to \lambda \mathbf{x^i}$ which is continuous (multiplication in reals).

That is, the collection of continuous maps from any topological space $X$ to $\mathbb{R}^m$ form a vector space over $\mathbb{R}$, with pointwise addition and scalar multiplication.

7 - Sequences

a map $f: \mathbb{N} \to X$ where $X$ is a metric space is called a sequence.

we say $f$ is cauchy if its terms get arbitrary close, taking a far enough tail. That is, for each $\epsilon > 0$, there exists $N \in \mathbb{N}$ for which whenever $i,j > N$, we have $d(f_{i},f_{j}) < \epsilon$.

We will denote the $n$-tail of a sequence by $f_{>n}$. A subsequence of $f$ is obtained by selecting terms along a strictly increasing map $A: \mathbb{N} \to \mathbb{N}$. It is given by $f \circ A$, and will be denoted $f|A$.

we say for some $p \in X$, $f$ converges to $p$ and write $f \longrightarrow p$ when for each $\epsilon > 0$, there exists $N \in \mathbb{N}$ for which $f_{>N} \subset B_{\epsilon}(p)$, that is a tail that is $\epsilon$-close to $p$.

next we claim that the following two statements are equivalent:

1. $f: \mathbb N \to X$ converges to $p \in X$.
2. $f$ is cauchy and, $f$ is eventually constant at $p$ or $p$ is a (THE) limit point of $f(\mathbb N)$.

$1 \Rightarrow 2$: $f$ is cauchy: we know that a tail of $f$ is $\epsilon/2$ close to $p$ for any $\epsilon>0$, thus by the triangle inequality, that same tail is $\epsilon$ close to each other.

now, if $f$ is eventually constant at $p$ all balls (WITH CENTRE LOOK AT THE DEFINITION OF CONVERGENCE) at least contain the constant $p$-tail.

Now for any $\epsilon > 0$ some $N$-tail is in the open $\epsilon,p$ ball (with centre included), since the sequence is not eventually constant at $p$ there is an infinite subsequence of this tail contained inside the open $\epsilon,p$ ball that are NOT equal to $p$! thus the ball without centre has a nonempty intersection with $f(\mathbb N)$ thus $p$ is a (THE) limit point of $f(\mathbb N)$.

$2 \Rightarrow 1$: I mean if something is eventually constant at $p$ then of course it converges to $p$.

suppose $p$ is a limit point of $f(\mathbb N)$, the intuition we have is that $f$ is cauchy, meaning tails are eventually $\epsilon$ close to each other, and further that for any $\epsilon > 0$ some subsequence (infinite ofc) is $\epsilon$ close to $p$.

somehow we want to put these together to show that tails are $\epsilon$ close to $p$.

well, suppose there is a pathological $\epsilon^*$ for which no tail is fully contained inside the $p,\epsilon^*$ ball, that is to say, the set of $n$ such that $d(f(n),p) \ge \epsilon^*$ has no maximum element. (if it did, you could take the tail beyond that point)

similarly, given that $p$ is a limit point, the set of $n$ for which $d(f(n),p) < \epsilon^*/2$ has no maximum element either.

that is to say, in the tail that is $\epsilon^*/2$ close to each other, you will find both kinds of elements and then you can take the triangle inequality to show that that shit is a contradiction.

further you can show that if $f \to p$ and $f \to q$ then $p = q$.

suppose $p \neq q$. take $d(p,q)/2$ balls around $p$ and $q$. a tail is fully contained in one ball, and a tail is fully contained in the other ball. well one of them is a subtail of the other one, so we have a tail contained in two disjoint sets, which doesn’t make sense.

Convergent Subsequences in Compact Spaces

Now, we claim that if $f: \mathbb{N} \to X$ is a sequence into a compact metric space, then it has a convergent subsequence.

Let $J_{-1} = \mathbb{N}$. Take $\epsilon_k = 1/2^{k+2}$. Cover $X$ by open $\epsilon_k$ balls and take a finite subcover. Given that $J_{k-1}$ has no maximum element, some ball $C_k$ in this finite subcover has the property that

$$J_k = J_{k-1} \cap f^{-1}(C_k)$$

has no maximum element. Indeed, if each of these sets had a maximum, then their finite union $J_{k-1}$ would have a maximum.

Thus, by induction, we have a sequence of nested sets

$$J_0 \supset J_1 \supset J_2 \supset \dots$$

such that $J_k$ has no maximum element and, whenever $i,j \in J_k$,

$$d(f_i,f_j) < 2\epsilon_k = \frac{1}{2^{k+1}}.$$

Pick $j_0 \in J_0$. Having picked $j_0 < j_1 < \dots < j_k$ with $j_i \in J_i$, we can pick $j_{k+1} \in J_{k+1}$ such that $j_{k+1} > j_k$, since $J_{k+1}$ has no maximum element.

Let $A: \mathbb{N} \to \mathbb{N}$ be given by $A(k)=j_k$. We claim that the subsequence $f \circ A$ is cauchy. For any $\epsilon>0$, take $K$ such that $1/2^{K+1}<\epsilon$. If $m,n \ge K$, then $j_m,j_n \in J_K$, and hence

$$d(f_{j_m},f_{j_n}) < \frac{1}{2^{K+1}} < \epsilon.$$

If $(f \circ A)(\mathbb{N})$ is finite, then $f \circ A$ is eventually constant. This is immediate if its image has one point. Otherwise, take $\delta>0$ smaller than the minimum distance between two distinct points of its image. A $\delta$-cauchy tail can contain only one point.

Suppose instead that $(f \circ A)(\mathbb{N})$ is infinite. We claim that it has a limit point in $X$. Otherwise, for each $x \in X$, there is an open ball $B_x$ such that

$$(B_x-\{x\}) \cap (f \circ A)(\mathbb{N}) = \varnothing.$$

The balls $B_x$ cover $X$, so take a finite subcover. Each ball in this finite subcover contains at most one point of $(f \circ A)(\mathbb{N})$, which would make $(f \circ A)(\mathbb{N})$ finite, a contradiction.

Thus $f \circ A$ is cauchy and is either eventually constant or has a limit point. By the equivalence proved above, it converges.