# A multiple choice question on topology regrading continuity of identity function with different topologies.

Let $X$ be a set with at least two elements. Let $\tau, \tau'$ be two topologies on $X$ such that $\tau'\neq\{\emptyset,X\}$. Which of the following condition is necessary for the identity function to be continuous.

$id: (X,\tau)\rightarrow (X,\tau')$

(1) $\tau \subseteq \tau'$

(2) $\tau'\subseteq\tau$

(3) no condition on $\tau$ and $\tau'$

(4) $\tau \cap \tau'=\{\emptyset, X\}$

My Efforts

(1) False

Let $U$ be an open set in $\tau'$, that is not in $\tau$. Inverse image of $U$ is $U$ but the preimage is not open.

(2) True.

Every open set in $\tau'$ is already open in $\tau$ so identity function is continuous.

How should I approve or reject other two option.

What if the topologies are not comparable;?

• Please notice the difference between \phi and \emptyset. – tst Sep 9 '18 at 6:53
• You have already identified the correct answer. – Lord Shark the Unknown Sep 9 '18 at 6:56
• @LordSharktheUnknown What if the topologies are not comparable, how do I show that result in option 2 is actually an if and only if statement – StammeringMathematician Sep 9 '18 at 7:07
• You can prove 2. is an if and only if by applying the definition of continuity @StammeringMathematician – Lord Shark the Unknown Sep 9 '18 at 7:09

Given two topologies $\tau,\tau'$ on the same set $X$, the identity function $id:(X,\tau)\to(X,\tau')$ is continuous exactly when $\tau'\subseteq\tau$ for the reason you gave. This forces comparability between the topologies. Thus, the identity function is never continuous when $\tau$ and $\tau'$ are not comparable.
As for why $4)$ is incorrect, think about the continuity condition. We want for every set $U\in\tau'$ that $U\in\tau$. However, if $\tau\cap\tau' = \{\emptyset, X\}$, then this condition fails on every nontrivial open set. It is (in a sense) the 'worst case scenario' for continuity.