Are there powerful ways to use the topological definition of continuity in real analysis?

Question

In the lectures for introductory real analysis, my professor repeatedly told the class that the topological definition of continuity (preimage of open is open) is the most powerful version of continuity, most useful...

Then the only example he came up with was ....

Example: Prove $S = \{x| ``{\text{weird, impractical and obviously ad hoc inequality here}}"\}$ is open

Then the LHS of this weird inequality is a continuous function and inequality is an open set and the preimage $f^{-1}$ on that open set is open, hence $S$ is open.

Then we never discussed this again and it never came up in any other context as far I can remember.

Question, are there more practical examples/ways to use the topological version of continuity in real analysis?

Answer 1

Calling it the "most powerful" is obviously wrong, as the pointwise definition of continuity implies the "preimage of open is open" definition for global continuity, while also allowing you to discuss functions that are not globally continuous.

But for globally continuous functions, it is a very useful characterization. For example:

Theorem: The continuous image of a compact set is compact.

Proof: Let $f : X \to Y$ and let $A \subset X$ be compact. if $\mathscr S$ is an open cover of $f(A)$, then $\{f^{-1}(S)\mid S \in \mathscr S\}$ is an open cover of $A$, and so must have a finite subcover $\{f^{-1}(S_i)\}_{i=0}^n$. But then $\{S_i\}_{i=0}^n$ must be a subcover of $f(A)$. Therefore $f(A)$ is compact.

Theorem: the continuous image of a connected set is connected.

Proof: Let $f : X \to Y$ and let $A \subset X$. Suppose $f(A)$ is disconnected. Then there are open sets $U, V$ in $Y$ such that $U \cap f(A) \ne \emptyset, V \cap f(A) \ne \emptyset, U \cap V = \emptyset$ and $ f(A) \subset U\cup V$. But then $f^{-1}(U) \cap A \ne \emptyset, f^{-1}(V) \cap A \ne \emptyset, f^{-1}(U) \cap f^{-1}(V) = \emptyset$ and $A \subset f^{-1}(U)\cup f^{-1}(V)$. Since $f^{-1}(U)$ and $f^{-1}(V)$ are open, they form a disconnection of $A$. By the contrapositive, if $A$ is connected, the so is $f(A)$.

There are two very powerful results proven easily through the "topological" definition of continuity. But the argument is strictly topology. So what about analysis? Recall the Heine-Borel theorem: a subset of $\Bbb R$ is compact if and only if it is closed and bounded. And recall that a subset of $\Bbb R$ is connected if and only if it is an interval.

The two results above then have two easy corollaries:

Extreme Value Theorem: If $f$ is continuous on the closed interval $[a,b]$, then there exists $c, d \in [a, b]$ such that for all $x \in [a,b], f(c) \le f(x) \le f(d)$.

Proof: $f([a, b])$ is compact and connected, and therefore is a closed and bounded interval: $f([a,b]) = [m,M]$. Hence there are $c, d \in [a,b]$ with $f(c) = m$ and $f(d) = M$, and for any $x\in[a,b], f(x) \in [m,M]$, and so $f(c) \le f(x) \le f(d)$.

Intermediate Value Theorem: If $f$ is continuous on the closed interval $[a,b]$ and either $f(a) < c < f(b)$ or $f(b) < c < f(a)$, then there is an $x \in (a,b)$ such that $f(x) = c$.

Proof: As with the Extreme Value Theorem, $f([a,b])$ is an interval. Therefore every element of that interval is the image of some point in $[a,b]$.

These are two of the most powerful results in real analysis. And they follow easily from the Heine-Borel theorem when one uses the "topological" definition of continuity. Proving them from the epsilon-delta definition is significantly more involved.

Answer 2

Here is a very simple example. If $f:\mathbb{R^n}\to \mathbb{R}$ is a continuous function, and $c \in \mathbb{R}$, then $f^{-1}(c)$ is a closed set. For instance, the graph of the hyperbola $xy = 1$, that is the set $\{(x,y) \in \mathbb{R}^2 \mid xy = 1 \}$, is closed. Similarly, the circle $x^2 + y^2 = 1$ or more generally the set of zeroes of a polynomial function, is closed.