Analysis Problem of the Day 12

The following somewhat tricky problem appeared as Problem 5 on Stanford’s Fall 2023 Analysis Qual:

Problem 5: Let $E \subset \mathbb{R}^n$ and let $f: E \to \mathbb{R}^n.$ Define

$F:=\{x \in E: \exists \{x_k\}_{k=1}^\infty \subset E \setminus \{x\}, x_k \to x, f(x_k) \to f(x)\}.$

Show that $E \setminus F$ is at most countable.

Solution: This problem reminds us of the sequential definition of continuity of a function at a point, namely, that $f$ is continuous at $x$ if for all sequences $x_n \to x,$ one has $f(x_n) \to f(x).$ However, the crucial difference here is that $F$ is the set of points not where every sequence satisfies this property, but where at least one such sequence exists. Thus, we would expect $F$ to be much larger than the set of points of continuity of $E$ (which, by the way, is a dense $G_\delta$ subset of $E.$ )

Motivated by this, one might wonder if there is an easier way to rephrase this definition, as for the definition of continuity at a point, using $\epsilon-\delta$ arguments. In fact, there is: define

$C:=\{x \in E: \forall \epsilon >0, \forall \delta>0, \exists y \in B_\delta(x): f(y) \in B_\epsilon(f(x))\}.$

Note that if $x \in F,$ then for all $\epsilon,\delta>0,$ there exists an $x_k$ such that $x_k \in B_\delta(x)$ and $f(x_k) \in B_\epsilon(f(x))$ (the first condition can be satisfied since $x_k \to x$ and the second since $f(x_k) \to f(x).$ ) Conversely, if $x \in C,$ setting $\epsilon = \delta = \frac{1}{n}$ yields a sequence $x_k \to x$ such that $f(x_k) \to f(x).$ Thus $F=C.$

The advantage of using this new set $C$ is that it is much easier to describe its complement. Namely,

$E \setminus C = \{x \in E: \exists \epsilon >0, \exists \delta > 0, f(B_\delta(x)\setminus\{x\}) \cap B_\epsilon(f(x)) = \varnothing\}.$

In other words, $E \setminus C$ is the set of points $x$ where the image of $f$ around $x$ is away from $f(x).$

Another key point here will be played by isolated values – this is where we need a classic fact from topology that a second-countable space (in particular, any subspace of $\mathbb{R}^n$ ) has at most countably many isolated values. The proof of this is straightforward – since $\{x\}$ is an open set, it contains a basis element, so $\{x\}$ is itself a basis element, and since there are at most countably many basis elements, there are at most countably many isolated ones. Indeed, if this were not the case, the set of isolated values of $E$ would be in $E \setminus F$ (since no sequence $x_k$ converging to any such element would exist) and be uncountable, contradicting what we are being asked to prove. Thus, a crucial fact in the proof will be the second-countability of the space, namely, that there exists a countable topological base. This motivates the following approach.

I claim that $E \setminus F$ is the union of all isolated values of $f^{-1}(U),$ where $U$ ranges over the countable topological basis of $f(E).$ Indeed, if $x \in E \setminus F,$ then $f^{-1}(B_\epsilon(f(x))) \cap B_\delta(x) = \{x\},$ so for some $U \subset B_\epsilon(f(x)),$ $x$ is an isolated value of $f^{-1}(U).$ Conversely, if $x$ is an isolated value of $f^{-1}(U)$ for some basis set $U,$ then for some $\delta>0,$ $f(B_\delta(x)\setminus\{x\}) \cap U = \varnothing,$ so there exists $\epsilon > 0$ such that $f(B_\delta(x) \setminus \{x\}) \cap B_\epsilon(f(x)) = \varnothing.$

Finally, since $f^{-1}(U)$ is a subspace of a second-countable space, the isolated values of $f^{-1}(U)$ form at most a countable set, so $E \setminus F$ is at most a countable set as a countable union of at most countable sets.

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