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LetX={g,s}, and endow X with the following topology: The subsets{∅,X,{g}} are open. Give[0,1]the usual metric topology. (a) Suppose f:X→ [0,1] is a continuous function such that f(s)=0. Show that f(g)=0.

Ok I am probably misunderstanding something, but the singleton {0} is a closed set in [0,1] right? Then its preimage must also be closed but {g} is open in (X,T). How can that be?

(b) Produce, with proof, a nonconstant continuous function f:[0,1] → X.

Since X only has two elements, I can only map a part of the interval of [0,1] to g and the remaining part to s. Now one part of this has to be an interval to be open and the other part should be closed, so either a closed interval or a singleton. So I was thinking something like [0.1) → g and {1} → s. The preimage of g is open and the preimage of s is closed so this should be a continuous map.

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  • $\begingroup$ A set can be open and closed at the same time. $\endgroup$
    – Henno Brandsma
    Apr 22, 2021 at 11:41

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$f(s)=0$ means that $s \in f^{-1}[[0,\frac1n)]$ and the latter set is open (for any $n \in \Bbb N$) as $[0,\frac1n)$ is open in $[0,1]$ and $f$ is continuous. The only open set in $X$ that contains $s$ is $X$ so in fact $X=f^{-1}[[0,\frac1n)]$, so $f(g) < \frac1n$ for all $n$ which implies $f(g)=0$. QED.

For (b) your example works The proof is just noting that $f^{-1}[\{g\}]$ is open (and $\{g\}$ is the only non-trivial open set in $X$). No words needed on $s$ or closed sets etc.

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  • $\begingroup$ Hi. Based on the comment you left, it it implying that {0} is a clopen set of [0,1]? But how can that be ? There is no open ball that can fit inside 0, $\endgroup$
    – willyx888
    Apr 22, 2021 at 12:21
  • $\begingroup$ @willyx888 you said the preimage bust be closed but $\{g\}$ is open. That $\{g\}$ is open is not the issue, it's that $\{g\}$ is not closed. In $[0,1]$ $\{0\}$ is closed but not open. But it could have an open pre-image in general. $\endgroup$
    – Henno Brandsma
    Apr 22, 2021 at 12:28
  • $\begingroup$ The function is continuous so does that not imply the perimage of open sets must be open and preimage of closed sets must be closed? I thought that is the definition of continuity. I understand the proof as a whole. The preimage of {0} is the entire space which is also considered closed. $\endgroup$
    – willyx888
    Apr 22, 2021 at 12:35
  • $\begingroup$ @willyx888 Yes, but still the pre-image of a closed set can also be open and the pre-image of an open set can also be closed, also for continuous $f$. In your remark, $\{g\}$ is open does not mean (in general) that $\{g\}$ is not closed. So as an argument it does not work. $\endgroup$
    – Henno Brandsma
    Apr 22, 2021 at 12:36

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