# Given the set X={g,s} with the following topology , show that the continuous function f(g)=0

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.

• A set can be open and closed at the same time. Apr 22, 2021 at 11:41

$$f(s)=0$$ 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.

• 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, Apr 22, 2021 at 12:21
• @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. Apr 22, 2021 at 12:28
• 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. Apr 22, 2021 at 12:35
• @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. Apr 22, 2021 at 12:36