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.


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

Thanks in advance!

  • $\begingroup$ $f:(X,\tau_1)\to(X,\tau_2)$ is (by definition) continuous if for every $B\in\tau_2$ we have: $f^{-1}(B)\in\tau_1$. $\endgroup$ – drhab Nov 3 '18 at 15:22
  • $\begingroup$ @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 $\implies$. $\endgroup$ – Pedro Gomes Nov 3 '18 at 15:25
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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 $f(U)\subseteq V$ $\endgroup$ – Pedro Gomes Nov 3 '18 at 15:17
  • $\begingroup$ 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 $U$. $\endgroup$ – b00n heT Nov 3 '18 at 15:20
  • $\begingroup$ With the definition you provided my proof $\implies$ would be right? $\endgroup$ – Pedro Gomes Nov 3 '18 at 15:21
  • 1
    $\begingroup$ 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 $\endgroup$ – b00n heT Nov 3 '18 at 15:28

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