## 247 Review Post #2

Ok so, I’m back to talk about some more lovely things that happen in $\mathbb{R}^n$. This is one of the things I feel is neglected far too much by all but the most rigorous calculus courses: sequences are really fundamentally intertwined with all of the underpinnings of why calculus (in one variable, and also in several) works. It seems that so much just reduces to something involving sequences.

Take topology in $\mathbb{R}^n$ for example. You start with the notion of an interior point, and from there you define an open set to be a set consisting only of interior points. Often after this, we proceed to define a closed set as “a set whose complement is an open set”. This is one characterization of the property of “closedness”. An alternative characterization you may see is that closed sets contain their closures.

The closure of a set $A$ is a (usually larger) set which consists precisely of all the points $\vec{x}$ in the space such that some sequence in $A$ converges to $\vec{x}$. If $A$ is closed, however, one can show that all of the points in the closure also lie inside $A$ itself, and vice versa (it’s an equivalence).

That said, when we talk about compact sets (sets that are both closed and bounded), some rather beautiful interplay occurs between the structure of these sets and the nature of continuous functions. An alternative characterization of continuity is the classic “if a function is continuous at a point, then the limit of the images of a sequence converging to that point is the image of the limit” (ie the point itself). With this in hand, I’ll show something that makes the EVT a one-line proof.

Theorem. Let $A \subseteq \mathbb{R}^n$ be a compact set, and let $f : A \to \mathbb{R}^m$ be a continuous function. Then the image set, $f(A)$, is compact as well.
Proof. First, we denote $B := f(A)$. We want to show that $B$ is closed. Note that $\mathrm{cl}(B) \supseteq B$ always, so we need only show $\mathrm{cl}(B) \subseteq B$. To do this, let us choose an arbitrary $\vec{y}_0 \in \mathrm{cl}(B)$. By definition of the closure of $B$, there exists some sequence $(\vec{y}_k)_{k=1}^\infty$ in $B$ such that $\vec{y}_k \to \vec{y}_0$. But surely since each $\vec{y}_k \in B$, there exists a corresponding sequence of points $(\vec{x}_k)_{k=1}^\infty$ in $A$ such that $f(\vec{x}_k) = \vec{y}_k$. We now simply apply the Bolzano-Weierstrass theorem for $\mathbb{R}^n$, which tells us that since $A$ is compact, every sequence in $A$ has a subsequence converging to a limit which lies within $A$. Hence we get the subsequence $(\vec{x}_{k(p)})_{p=1}^\infty$ such that $\vec{x}_{k(p)} \to \vec{x}_0$ for some $\vec{x}_0 \in A$. Now, the continuity of $f$ gives us that $\vec{y}_{k(p)} = f(\vec{x}_{k(p)}) \to f(\vec{x}_0)$. But we already know that $\vec{y}_k \to \vec{y}_0$. So by the uniqueness of the limit, $\vec{y}_0 = f(\vec{x}_0)$! Hence $\vec{y}_0 \in B$, so $B$ is closed.

We now must show that $B$ is bounded. This is done in much the same way: we assume otherwise. Then for each natural $k$, we obtain a point $\vec{y}_k$ such that $\| \vec{y}_k \| \geq k$. Now again obtain the corresponding sequence $(\vec{x}_k)_{k=1}^\infty$ in $A$, choose a convergent subsequence, use the continuity of $f$, and we quickly obtain a convergent sequence that is unbounded. This is surely a contradiction since every convergent sequence is bounded. Hence $B$ must be bounded, and finally this allows us to conclude that $B$ is compact.

OK, I think that’s good enough for the basics. Now to understand the 3-page proof that any function continuous modulo a null set is integrable… which draws on null sets (duh), compactness, and uniform continuity. The secret is
$\epsilon_0 = \frac{\epsilon}{(\beta - \alpha) + \mathrm{vol}(P) + 1}$!