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Could you please verify this (rather simple) proof? I'm a bit new to this kind of reasoning.

The question is:

Let $\mathcal{T}_1$, $\mathcal{T}_2$ be two topologies on some set $X$, when is the identity map $id:X \to X$, $id(x) = x$ continuous?

Here is my reasoning:

If $\mathcal{T}_1 = \mathcal{T}_2$, then for any $S \in \mathcal{T}_2$, $id^{-1} (S) = S \in \mathcal{T}_1$ so then it's continuous.

If $\mathcal{T}_2 \subseteq \mathcal{T}_1$, then it's exactly like above, then it's also continouos.

Now:

If $\mathcal{T}_1 \subseteq \mathcal{T}_2$ then take some $x \in U \in \mathcal{T}_2 \setminus \mathcal{T}_1$, then $id^{-1}(U) \not\in \mathcal{T}_1$, so it's not continuous.

So the identity map is continuous if $\mathcal{T}_1$ is finer than $\mathcal{T}_2$.

Is this OK?

Is there a better way to show this?

Any extra applicable theory?

Thanks in advance

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  • $\begingroup$ I didn't get what is problem in proof but I like it. It helped a lot. $\endgroup$ Nov 9, 2017 at 14:38

2 Answers 2

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That's almost perfect. Rather, you've (almost) showed that it's continuous if and only if $\mathcal T_1$ is finer than $\mathcal T_2.$

You only need two cases: $\mathcal T_2\subseteq\mathcal T_1,$ and $\mathcal T_2\not\subseteq\mathcal T_1.$ In the former case, you'll reason as you did in your equality case. In the latter case, you'll reason as you did in your $\mathcal T_1\subseteq\mathcal T_2$ approach, but it will actually be viable reasoning. (Can you figure out why yours was not?)

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  • $\begingroup$ I'm thinking that I messed up the subset inclusions, im going from $(X, \mathcal{T}_1) $ to $(X, \mathcal{T}_2) $, so if $\mathcal{T}_1 \subset \mathcal{T}_2$, then $id$ isn't continuous, and if $\mathcal{T}_2 \subseteq \mathcal{T}_1$, then it is continuous. Is this better? $\endgroup$
    – JuliusL33t
    Mar 29, 2015 at 17:02
  • $\begingroup$ @JuliusL33t you still don't cover all possibilities. It is enough that $\mathcal{T}_2 ⊈ \mathcal{T}_1$ to $f$ not be continuous. $\endgroup$
    – user87690
    Mar 29, 2015 at 19:08
  • $\begingroup$ @JuliusL33t: I was the one who reversed my inclusions. Oops! (Fixed now.) As user87690 points out, it isn't enough to show what happens when $\mathcal T_1⊊\mathcal T_2$, since the two topologies needn't be comparable. The mistake I was referring to was that you actually allowed equality in your third case, so there might be no such $U$! (Also, there is no need to refer to $x$, but that's more a simplicity issue than an error in reasoning.) $\endgroup$ Mar 30, 2015 at 12:03
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Identity map $id:(X,\tau_1)\to X,(\tau_2)$ is continuous if and only if $\tau_2\subseteq\tau_1$.

Continuity of $id$ function is not necessary if $\tau_1\subseteq\tau_2$.

For example:

$id:(X,\tau_{\text{trivial topology}})\to(X,\tau_{\text{discrete topology}})$

Now look at the inverse image of open sets.

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