If $f:X \to \mathbb{R}$ is continuous, then f(x) is a constant function

Question

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Proposition: Let $(X,S)$ and $(Y,T)$ be topological spaces. Suppose $f: X \to Y$ is a continuous function. Let

$X = \{A, B, C \}$,

$S = \{ \emptyset, \{A\}, \{A, B \}, \{A,B,C\} \}$,

$Y = \mathbb{R}$, satisdying the standard topology defintion.

Show that if $f$ is continuous then it must be a constant function; i.e. $f(x) = C$, for some $C \in \mathbb{R}$

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I understand the topological def. of continuity. I also understand how to prove the converse. If f is constant, then $ \forall t \in T , f^{-1}(t)$ is either $ \emptyset$ or entire $X$, and both are open in $(X,S)$. Hence $f$ is continuous.

But to prove this statement, I am not sure how to go about it. I am thinking of two ways,

1) Contradiction. Assuming $y,z \in T$ and $y \neq z$. Not sure how to proceed.

2) Unpacking the def. of topological continuity and showing that only $f^{-1}(\emptyset)$ and $f^{-1}(T)$ are the only elements that are open in $(X,S)$.

Answer 1

Hint. You can show that $f^{-1}[\{f(A)\}]$ is a closed set containing $A$. Which closed set can contain $A$?

Answer 2

Let $Y$ be any Hausdorff space. If $f(X,\mathcal{S}) \to Y$ is continuous and non-constant then let $y_1 \neq y_2$ be two points in $f[X]$. Let $U$ and $B$ be disjoint open neighbourhoods of $y_1$ resp. $y_2$ by Hausdorffness.

Then $f^{-1}[U]$ and $f^{-1}[V]$ are open and non-empty and $f^{-1}[U] \cap f^{-1}[V]= f^{-1}[U \cap V]=\emptyset$ but in $(X,\mathcal{S}$ any two non-empty open sets intersect (in $A$. Contradiction, so $f$ cannot be non-constant.

Note that this applies to your $Y$ as we can take $U=(\leftarrow, z)$ and $V=(z,\rightarrow)$ when $y_1 < z < y_2$ in $\Bbb R$ (or $y_2 < z< y_1$ and then reverse $U$ and $V$).