Proving that iff the preimage of a closed set is closed, then the function is continuous

Question

I am trying to prove that iff the preimage of a closed set is closed, then the function is continuous, without taking the set complement (instead, I am supposed to be using limit points). I have already proven this by first proving the preimage of an open set is open, but this was an additional problem in my homework. This problem is just restricted to real vector spaces.

Here is my proof so far.

Suppose that $f:\mathbb{R}^m \rightarrow \mathbb{R}^n$ is continuous on $ \mathbb{R}^m $ and $ U $ is a closed set in $ \mathbb{R}^n $. Let $\textbf{x}$ be a limit point of $ f^{-1}(U) $. Since $f$ is continuous, then $f(\textbf{x})$ is a limit point of $U$.

(Here I am taking the definition of continuity that for any sequence $x_k \rightarrow x$, we will have $f(x_k) \rightarrow f(x)$.)

Since $f(\textbf{x}) \in U$ as $U$ is closed, then $\textbf{x} \in f^{-1}(U)$. So $f^{-1}(U)$ contains all of its limit points, and is therefore closed.

Now suppose instead that $U$ is a closed set in $ \mathbb{R}^n$ and $f^{-1}(U)$ is a closed set in $\mathbb{R}^m$.

Here is where I get stuck. I can't see how I can use the fact that $U$ and $f^{-1}(U)$ contain all their limit points to prove that $f$ must be continuous. I have tried to start with something along the lines of "Suppose that x is a limit point, so $\textbf{x}_k\rightarrow \textbf{x}$, but I can't seem to get the seemingly trivial step of proving that $f(\textbf{x}_k) \rightarrow f(\textbf{x})$. I am almost certain I have missed something obvious. Any help would be appreciated.

Answer 1

A correction: the hypothesis for the other implication should read:

Now suppose that whenever $U$ is a closed set in $\mathbb{R}^n$, $f^{-1}(U)$ is a closed set in $\mathbb{R}^m$.

To prove this implication, you can show the contrapositive. Assume that $f$ is not continuous, so it is not continuous at some point $x_0 \in \mathbb{R}^m$. Then, we will find a closed set $U \subseteq \mathbb{R}^n$ such that $f^{-1}(U)$ is not a closed subset of $\mathbb{R}^m$.

Take a sequence $\{ x_n \}_{n \in \mathbb{N}}$ in $\mathbb{R}^m$ converging to $x_0$ such that $\{ f(x_n) \}_{n \in \mathbb{N}}$ does not converge to $f(x_0)$ in $\mathbb{R}^n$. This could mean one of two things: either $f(x_0)$ does not lie in the closure of $A = \{ f(x_n) : n \in \mathbb{N} \}$, or $f(x_0)$ does lie in the closure of $A$ but $f(x_n)$ does not converge to $f(x_0)$.

The first case is easy to dispose of: $f(x_0) \not\in cl(A) \implies x_0 \not\in f^{-1}(cl(A))$. But $x_0$ lies in the closure of $B = \{ x_n : n \in \mathbb{N} \}$, and $B \subseteq f^{-1}(cl(A))$, so we have a contradiction.

For the second case, if $f(x_n)$ does not converge to $f(x_0)$, then there exists a subsequence $\{ f(x_{n_k}) \}_{k \in \mathbb{N}}$ such that every term is "far away" from $f(x_0)$. That is, there exists $\epsilon_0 > 0$ such that $|f(x_{n_k}) - f(x_0)| \geq \epsilon_0$ for all $k \in \mathbb{N}$. Let $A_1 = \{ f(x_{n_k}) : k \in \mathbb{N} \}$. Then, $\{ x_{n_k} \}_{k \in \mathbb{N}}$ is a sequence in $\mathbb{R}^m$ converging to $x_0$, but $f(x_0)$ does not lie in the closure of the set $A_1$. This is the case dealt with previously, so again we are done.

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Note that @karmalu's answer is incorrect by the following example. Consider the function $f : \mathbb{R} \to \mathbb{R}$ given by $$ f(x) = \begin{cases} 0, & x < 0;\\ 1, & x \geq 0. \end{cases} $$ Choose the sequence $\{ x_n \}_{n \in \mathbb{N}}$ given by $$ x_n = \frac{(-1)^n}{n}. $$ Then, $x_n$ converges to the point $x_0 = 0$. Moreover, $f(x_0) = 1$ lies in the closure of $\{ f(x_n) \}_{n \in \mathbb{N}_{>i}}$ for each $i \in \mathbb{N}$, because $$ f(x_n) = \begin{cases} 0, & n \mathrm{\ odd};\\ 1, & n \mathrm{\ even}. \end{cases} $$ Yet, $1$ is not the limit of $\{ f(x_{n}) \}_{n \in \mathbb{N}}$ because this is not a convergent sequence.

Answer 2

Take the closures of ${f(x_n)}_{n\in \mathbb{N}_{>i}} $ for increasing $i$. You have that $f(x) $ is in the closures for every $i$ because $x$ is in the pre-image. As this is true for every subsuccession $f(x) $ is the limit of the succession.