# Proving the identity functions is continuous iff $\tau_1\supseteq\tau_2$

Exercise: Let $$\tau_1$$ and $$\tau_2$$ be two topological spaces on a set $$X$$. Then $$\tau_1$$ is said to be a finer topology than $$\tau_2$$ if $$\tau_1\supseteq\tau_2$$.

Prove the identity function $$f:(X,\tau_1)\to (X,\tau_2)$$ is continuous if and only if $$\tau_1$$ is a finer topology than $$\tau_2$$.

Attempted proof:

Let $$A\in\tau_1$$. If $$f$$ is continuous then $$\exists B\in\tau_2$$ such $$f(A)\subset B\implies A\subset B$$ for any $$A\in\tau_1$$ such that all open sets of $$\tau_1$$ are open sets of $$\tau_2$$ hence $$\tau_1\supseteq \tau_2$$.

If $$\tau_1$$ is a finer topology than $$\tau_2$$, for any $$B\in\tau_2\implies B\in\tau_1$$ which implies that $$f^{-1}(B)=B$$, then $$B\in\tau_1$$ which proves $$f$$ to be continuous.

Questions:

Is my proof right? If not. What is wrong? What are alternatives?

• $f:(X,\tau_1)\to(X,\tau_2)$ is (by definition) continuous if for every $B\in\tau_2$ we have: $f^{-1}(B)\in\tau_1$. – drhab Nov 3 '18 at 15:22
• @drhab If you are referring to the proof below. I am aware of that definition as I used it on $\leftarrow$. However I am worried about failing to understand what was the mistake in $\implies$. – Pedro Gomes Nov 3 '18 at 15:25
• The word "if" in my former comment was meant to be "iff". Then if you take the identity for function $f$ it almost immediately translates to what must be proved. – drhab Nov 4 '18 at 8:13

When working with continuity, you always have to use the pre-image, I don't understand what you are trying to prove for $$\implies$$.
The direction $$\Longleftarrow$$ is correct as you have shown it.
For $$\implies$$ I would have just done the following: for any open set $$B$$ in $$\tau_2$$ as $$f$$ is continuous, $$f^{-1}(B)\in \tau_1$$ (i.e. is open) by definition. But as $$f=id$$, $$f^{-1}(B)=B\in\tau_1$$ So $$\tau_2\subseteq \tau_1$$, as wanted.
• Thanks for your answer. For $\implies$ I used the definition of continuity. If $a\in X\:\exists U$ such that $a\in U$ then $f(a)\in V$ such that $f(U)\subseteq V$ – Pedro Gomes Nov 3 '18 at 15:17
• This is not the correct definition. The correct definition would be: for any $V\in\tau_2$ such that $f(x)\in V$ there exists a $U\in \tau_1$ such that $f(U)\subset V$. So basically $V$ comes first, not $U$. – b00n heT Nov 3 '18 at 15:20
• With the definition you provided my proof $\implies$ would be right? – Pedro Gomes Nov 3 '18 at 15:21
• You must be careful with the definition of finer: $\tau_1\supseteq \tau_2$ does not mean that for any open set $B$ in $\tau_2$ you can find an open set $A$ in $\tau_1$ contained in $B$, $A\subset B$. It means that the set $B$ itself is in $\tau_1$, which is different – b00n heT Nov 3 '18 at 15:28