# Prove that a function between metric spaces is continuous iff the preimage of any open set in the codomain space is an open set.

So, here's the full question:

Let $$f:(X,d) \to (Y,d')$$ be a function between two metric spaces. $$f$$ is continuous iff for each open set $$O \subseteq Y$$, $$f^{-1}(O)$$ is an open subset of $$X$$.

Proof Attempt:

Let $$f: (X,d) \to (Y,d')$$ be a function between two metric spaces and suppose that $$f$$ is continuous.

Let $$O$$ be an open set of $$Y$$. To prove that $$f^{-1}(O)$$ is open, we need to show that it is a neighbourhood of each of its points. Let $$p \in f^{-1}(O)$$ be arbitrary but fixed. Then, $$f(p) \in O$$. Since, $$O$$ is a neighbourhood of $$f(p)$$ and $$f$$ is continuous, $$f^{-1}(O)$$ is a neighbourhood of $$p$$ and, therefore, of each of its points. Hence, $$f^{-1}(O)$$ is an open subset of $$X$$.

Now, suppose that for each open set $$O$$ of $$Y$$, $$f^{-1}(O)$$ is an open subset of $$X$$. Let $$a \in X$$ be arbitrary but fixed. Let $$M$$ be a neighbourhood of $$f(a)$$. Then:

$$\exists \epsilon > 0: S'(f(a),\epsilon) \subseteq M$$

Since $$S'(f(a),\epsilon)$$ is an open set, we can see that $$f^{-1}(S'(f(a),\epsilon))$$ is an open set. So, it follows that it is a neighbourhood of $$a$$. In particular, since it is the case that:

$$f^{-1}(S'(f(a),\epsilon)) \subseteq f^{-1}(M)$$

it follows that $$f^{-1}(M)$$ is a neighbourhood of $$a$$. That proves that $$f$$ is continuous at $$a \in X$$ and, therefore, continuous.

Does the proof above work? If it doesn't, why? How can I fix it?

• what is your definition of continuous? – Exodd Jul 29 '20 at 23:22
• I'm using the neighbourhood formulation of continuity in metric spaces. – Abhi Jul 29 '20 at 23:24

Here I provide a slightly different way to prove the proposed result.

Let us start with the implication $$(\Rightarrow)$$ first.

Let $$\mathcal{O}\subseteq Y$$ be an open set. Then we have to prove that $$f^{-1}(\mathcal{O})$$ is open.

In order to do so, let's consider that $$x\in f^{-1}(\mathcal{O})$$.

Then $$f(x)\in\mathcal{O}$$. Since $$\mathcal{O}$$ is open, there exists an $$\varepsilon > 0$$ such that $$f(x)\in N_{\varepsilon}(f(x))\subseteq\mathcal{O}$$.

Consequently, due to the continuity of $$f$$, we conclude there is a $$\delta > 0$$ s.t. \begin{align*} y\in N_{\delta}(x) \Rightarrow f(y)\in N_{\varepsilon}(f(x))\subseteq\mathcal{O} \Rightarrow y\in f^{-1}(\mathcal{O}) \end{align*}

which proves that $$x\in N_{\delta}(x)\subseteq f^{-1}(\mathcal{O})$$, whence we conclude that $$f^{-1}(\mathcal{O})$$ is open.

We may now approach the second implication $$(\Leftarrow)$$.

We have to prove that for every $$\varepsilon > 0$$ and every $$x_{0}\in X$$, there corresponds a $$\delta > 0$$ s.t. for every $$x\in X$$ \begin{align*} d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f(x),f(x_{0})) < \varepsilon \end{align*}

Let $$x_{0}\in X$$ and $$\varepsilon > 0$$.

If we consider any open ball $$N_{\varepsilon}(f(x_{0}))\subseteq Y$$, we know that $$f^{-1}(N_{\varepsilon}(f(x_{0}))$$ is open due to the given assumption.

Since $$x_{0}\in f^{-1}(N_{\varepsilon}(f(x_{0}))$$, there exists an open ball s.t. $$x_{0}\in N_{\delta}(x_{0})\subseteq f^{-1}(N_{\varepsilon}(f(x_{0}))$$.

Finally, we conclude that for every $$x_{0}\in X$$ and every $$\varepsilon > 0$$, there corresponds a $$\delta > 0$$ s.t. \begin{align*} x\in N_{\delta}(x_{0}) \Rightarrow f(x)\in N_{\varepsilon}(f(x_{0})) \end{align*}

and we are done. Hopefully this helps.

• Ah yes, that corresponds somewhat closely to what I've written. I mean, obviously, mine is a little more ugly to look at since it's just words but still. Thank you, this helps :D – Abhi Jul 29 '20 at 23:26
• You are welcome. I am glad to be of help. – APCorreia Jul 29 '20 at 23:30