Real Analysis: Continuity
Continuity and Compactness
The study of topology so far has been fairly self contained, where concepts such as open sets and compactness have been presented and analyzed in their own right, but not connected to any other mathematical concepts, even ones as basic as functions. If the study of topology has left you asking "So what?" then this section is where the answers begin. This section connects topology to continuity, allowing us to derive significant theorems related to extrema of real valued functions and the topology of inverse functions. (In fact, the study of general topology may be regarded as the study of continuous functions and what properties are preserved by them.)
Continuity Preserves Compactness
Continuous functions preserve compactness. That is to say, the image of a compact set under a continuous function is itself compact. Formally, we can write the Continuity Preserves Compactness Theorem (CPC theorem) as follows: If $f$ is a continuous function that maps a compact metric space $X$ into a metric space $Y,$ then its image set $f(X)$ is compact. Note that the metric space $Y$ does not need to be compact (indeed, that's rather the point of the theorem). An immediate result of this is theorem is that if $X_0$ is a compact subset of $X,$ then $f(X_0)$ is also compact.
Extreme Value Theorem for $\mathbb{R}$
A further and more significant consequence of the CPC theorem is the Extreme Value Theorem for real-valued functions. Formally, if $f$ continuously maps a compact metric space $X$ into $\mathbb{R},$ then there exist two values $a, b \in X,$ such that $f(a) = \inf(f(X))$ and $f(b) = \sup(f(X)).$ Plainly, this means that $f$ attains its minimum and maximum at points in its domain.
For example, the extreme value theorem lets us conclude that any polynomial defined over an interval will contain its maximum. In contrast, it says no such this about polynomials defined over the entire real line. The polynomial may have a maximum or a minimum, such as $x^2$ or $-x^2$ do, but since $\mathbb{R}$ is not compact, we would need a different approach to inferring the existence of extrema.
A notable shortcoming of the extreme value theorem is that it does not give any information as to what the actual extrema might be or where in the domain they might exist. Furthermore, it does not give any hint as to how many values might be maxima or minima. It...only says that at least one of each have gotta be in there somewhere. To ascertain these values would require information about the function itself beyond its topological properties, which by design are agnostic as to its geometry.
Continuity, Compactness, and Inverses
Recall that if a function is bijective, then it has an inverse. If a bijective function is also continuous, we might imagine that its inverse is also continuous. Intuitively, this seems like it should be the case: since a continuous function is "smooth," and we can imagine the inverse as a swapping of the horizontal and vertical axes of its graph, then resulting graph seems like it would also be smooth. However, this is not always so!
Since continuity and bijectivity are not enough to imply a continuous inverse, we might ask - under what conditions will a continuous bijection have a continuous inverse? One such condition is the compactness of the domain. The continuous inverse theorem states that if $f$ is a continuous bijection and has a compact domain, then its inverse $f^{-1}$ is also continuous. Note that this theorem is only an implication, not a biconditional - a continuous bijection may have a continuous inverse even if its domain is not compact (the existence of which is a problem below), but this is not guaranteed.
Problems
Prove the CPC Theorem: If $f : X \rightarrow Y$ is continuous and $(X, d_X)$ is compact, then $f(X)$ is compact.
Let $\mathcal{S}$ be an open cover of $f(X).$ Since $f$ is continuous, for each $S_i \in \mathcal{S}$ we see that $f^{-1}(S_i)$ is open. Since $X$ is compact, there are finitely many set $S_i$ such that $X \subseteq f^{-1}(S_1) \cup \ldots \cup f^{-1}(S_n).$ Since $f(f^{-1}(T)) \subseteq T$ for all $T \subseteq Y,$ it follows that $f(X) \subseteq S_1 \cup \ldots \cup S_n.$ Therefore $f(X)$ is compact.
Extreme Value Theorem for $\mathbb{R}$: Consider $f : X \rightarrow \mathbb{R}.$ Show that if $f$ is continuous and $X$ is compact, then there exist two values $a, b \in X$ such that $f(a) = \inf(f(X))$ and $f(b) = \sup(f(X)).$
Let $f : X \rightarrow \mathbb{R}$ be continuous and $X$ be compact. By the CPC theorem $f(X)$ is compact. Since $f(X) \subseteq \mathbb{R},$ it follows by the Heine-Borel theorem that $f(X)$ is closed and bounded. Therefore $\inf(f(X)) \in f(X)$ and $\sup(f(X)) \in f(X).$ By definition of $f(X),$ it follows that there exist two values $a, b \in X$ such that $f(a) = \inf(f(X))$ and $f(b) = \sup(f(X)).$
Continuous Inverse Theorem: Show that if $f : X \rightarrow Y$ is continuous and bijective and $X$ is compact, then its inverse $f^{-1}$ is also continuous.
Recall that since $f$ is bijective, its inverse exists and is defined as $f^{-1}(f(x)) = x.$
Let $f : X \rightarrow Y$ be continuous and bijective, and let $X$ be compact. Let $S$ be an open set in $X.$ It follows that $S^c$ is closed. Since $S^c$ is closed and $X$ is compact, it follows that $S^c$ is also compact. Therefore $f(S^c)$ is compact. Because $f$ is bijective, $f(S) = f(S^c)^c.$ Therefore $f(S)$ is open. Since a function is continuous if and only if the preimage of every open set in its codomain is open in its domain, it follows that $f^{-1}$ is continuous.
Provide and example of a continuous bijection defined over a non-compact domain whose inverse is also continuous.
The identity function on the open interval $f : (0,1) \rightarrow (0,1)$ defined by $f(x) = x$ is a monomonial and therefore continuous. It is also a bijection whose inverse is $f^{-1}(x) = x.$ Since $f^{-1} = f,$ it follows that $f^{-1}$ is continuous, even though its domain is an open interval and therefore not compact.
Provide an example of a continuous bijection whose inverse is not continuous.
Consider the following function:
$f(x) = \left\{ \begin{array}{ll} x & 1 \leq x < 2\\x-1 & 3 < x \leq 4\end{array}\right.$
Note that $f$ is continuous and bijective. However, its inverse is defined as
$f^{-1}(x) = \left\{ \begin{array}{ll} x & 1 \leq x < 2\\x+1 & 2 < x \leq 3\end{array}\right.$
We can see that $f^{-1}$ is discontinuous when $x = 2.$