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.
Best Answer
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 $x\in\mathbb{R}^{m}$.