# $F \subseteq \mathbb{R}^n$, the set $f^{-1}(F)$ is a closed subset of $\mathbb{R}^m \implies f$ is continuous.

Let $$f:\mathbb{R}^m \to \mathbb{R}^n$$ be a function. The following are equivalent:

(1) f is continuous.

(2) For every closed set $$F \subseteq \mathbb{R}^n$$, the set $$f^{-1}(F) := \{ x \in \mathbb{R}^{m} : f(x) \in F \}$$ is a closed subset of $$\mathbb{R}^m$$.

How can I prove that $$(2) \implies (1)$$ using the following fact:

$$f$$ is continuous at $$a \in A \iff$$ For every sequence of points $$\{x_k\}_{k=1}^{\infty}$$ such that $$\lim_{k \to \infty} x_k = a$$ we have $$lim_{k \to \ \infty} f(x_k)= f(a)$$

I was thinking of doing a proof by contradiction:

suppose f is not continuous then

$$\exists_{a \in \mathbb{R}^m} \exists_{\epsilon > 0} \forall_{\delta > 0} \exists_{x \in \mathbb{R}^m} || a - x|| < \delta \text{ but } ||f(x) - f(a)|| \geq \epsilon$$

I know I can pick a sequence $$\{x_n\} \in \mathbb{R}^m$$ that converges to a, but I am not sure how to proceed.

• This is easy if you recall $f$ is continuous iff $f^{-1}(U)$ is open for all $U$ open.
– zhw.
Commented Mar 31, 2020 at 21:43

There are many different characterisations of continuity in metric spaces and therefore, also in $$\mathbb{R}^{n}$$. As zhw. mentioned in his/her comment, a map $$f$$ is continuous if and only if $$f^{-1}(\mathcal{O})$$ is open for every open set $$\mathcal{O}\in\mathbb{R}^{n}$$.

However, I think you are looking for a prove which uses the epsilon-delta-criterion, which states that $$f:\mathbb{R}^{m}\to\mathbb{R}^{n}$$ at is continuous $$x_{0}\in\mathbb{R}^{m}$$ if and only if

$$\forall\varepsilon\in\mathbb{R}_{>0}~\exists\delta\in\mathbb{R}_{>0}~\forall x\in\mathbb{R}^{m}: (\Vert x_{0}-x\Vert_{\mathbb{R}^{m}}<\delta\Rightarrow \Vert f(x_{0})-f(x)\Vert_{\mathbb{R}^{n}}<\varepsilon)$$

This is also basically the translation of your statement with the limits.

We assume (2) is true, which means that $$f^{-1}(A)$$ is closed for all closed subsets $$A\subset\mathbb{R}^{n}$$. Let $$x_{0}\in \mathbb{R}^{m}$$. Recall that $$A$$ is closed if and only if the complement $$\mathbb{R}^{n}$$\ $$A$$ is open.

We choose a closed set $$A\subset\mathbb{R}^{n}$$ with $$f(x_{0})\in\mathbb{R}^{n}$$\ $$A$$. Then there is a $$\varepsilon\in\mathbb{R}_{>0}$$ and an open ball $$B_{\varepsilon}(f(x_{0}))$$, such that $$f(x_{0})\in B_{\varepsilon}(f(x_{0}))\subset \mathbb{R}^{n}$$\ $$A$$. By assumption, $$f^{-1}(\mathbb{R}^{n}$$\ $$B_{\varepsilon}(f(x_{0}))=\mathbb{R}^{n}$$\ $$f^{-1}(B_{\varepsilon}(f(x_{0})))$$ is closed and therefore, $$f^{-1}(B_{\varepsilon}(f(x_{0})))$$ is open. Following this, there is a $$\delta\in\mathbb{R}_{>0}$$ and an open ball $$B_{\delta}(x_{0})$$ with $$x_{0}\in B_{\delta}(x_{0})\subset f^{-1}(B_{\varepsilon}(f(x_{0})))$$. This is exactly the epsilon-delta-criterion, when you plug in the definition of an open ball $$B_{\delta}(x_{0}):=\{x\in\mathbb{R}^{m}\mid \Vert x-x_{0}\Vert_{\mathbb{R}^{m}}<\delta\}.$$

Therefore, we have shown that $$f$$ is continuous at $$x_{0}$$ and since the choice of $$x_{0}$$ was arbitrary, $$f$$ is continuous at every point $$x\in\mathbb{R}^{m}$$.