3. Let $X$ and $X'$ denote a single set in the two topologies $\mathcal T$ and $\mathcal T'$, respectively. Let $i:X'\to X$ tbe the identity function.
(a) Show that $i$ is continuous $\Leftrightarrow \mathcal T'$ is finer than $\mathcal T$.
(b) Show that $i$ is a homeomorphism $\Leftrightarrow \mathcal{T'=T}$.

It's something about the first sentence that confuses me.
Okay, what I thought was: $X\in\mathcal T$ and $X'\in\mathcal T'$, where $X$ and $X'$ are (open) single sets, which I interpret as open sets having one element.

But I guess my interpretation is wrong, as the question doesn't make much sense in this way. So my question is how to interpret this question, and why I should interpret it this way. Don't spoil the answer!

  • $\begingroup$ Would you recommend this book? (I am looking to expand my topology books from 2 to a larger collection (like I have for Analysis)) $\endgroup$ – Alec Teal Mar 6 '14 at 22:35
  • $\begingroup$ @AlecTeal, yes. $\endgroup$ – msh210 Nov 29 '15 at 6:14

He means that $X$ and $X'$ are two topological spaces with the same underlying set. $X$ carries the topology $\mathcal T$ and $X'$ the topology $\mathcal T'$.

I would have formulated it in the following way:

Let $\mathcal T$ and $\mathcal T'$ be topologies on a set $X$ and consider the function \begin{align*} i\colon (X,\mathcal T')&\to (X,\mathcal T)\\ x&\mapsto x. \end{align*}

That way I would also have avoided to call $i$ the identity function; a true identity should always be continuos.

  • $\begingroup$ I think you are right. But how would you justify that statement in the question and your statement are equivalent ? $\endgroup$ – Kasper Mar 6 '14 at 16:45
  • $\begingroup$ I, too, was confused by the first sentence. I think that “in” is an unfortunate choice of preposition. But then I read on, and since the properties that are to be proved are standard observations, it was clear to me what must have been meant. I understand that this may be unsatisfactory. $\endgroup$ – Carsten S Mar 6 '14 at 18:51

What's meant is that the toplogies $\mathcal{T}$ and $\mathcal{T}'$ have a set, called both $X$ and $X'$ for some seemingly unnecessary reason, in common.

Regarding why it should be interpreted this way; well I guess that's just what it says in English.


I think, what is meant that you have two topological spaces $X = (X, \mathcal{T})$ and $X' = (X', \mathcal{T}')$ where $X=X'$ as sets, but not necessarily as topological spaces.

As for why: The question (b) means that it should be possible to make a statement about the equality of $\mathcal{T}$ and $\mathcal{T}'$ using "information" from $i$. Therefore $i$ should "use" the whole space, not only a subset. In other words if $i$ operated on a small subspace of the topolgical space, you would hardly be able to make a statement on the whole topology.

  • $\begingroup$ You are basicly saying, he must means something like this, otherwise it would be a very strange/confusing question. But if continue this line of reasoning. There are a LOT of question I find strange/confusing (at least the first time I read it). Should I freely interpret those question so that they become a lot easier :) :P ? $\endgroup$ – Kasper Mar 6 '14 at 16:36
  • $\begingroup$ Perhaps it would be clearer to write $X=(S,\mathcal T)$, $X'=(S,\mathcal T')$. $\endgroup$ – Carsten S Mar 6 '14 at 18:53
  • $\begingroup$ To me it seems not that "freely interpreted", in fact I tried to justify my interpretation after making it. @CarstenSchultz: Yes, but is very common to use the same symbol for the space and the set. Maybe a second goal of this question was to make aware of this, and that it can be confusing in this case. $\endgroup$ – Leonhard Mar 7 '14 at 9:45
  • $\begingroup$ That is a good point. $\endgroup$ – Carsten S Mar 7 '14 at 14:46

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