I have some confusion on the theorem: preimage of a closed set under a continuous function is closed. If we define $f:(0,1)\to \{0\}$ then $f$ is a continuous function since $f(x)=0$ for all $x$ in the domain, and we can apply the theorem so that $(0,1)$ is closed. But this is clearly not correct and I can't find what is wrong.
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1$\begingroup$ $(0,1)$ is a closed subset of $(0,1)$ (and more generally every topological space is open and closed in itself). It's just not closed in $\Bbb{R}$; that's a separate unrelated matter which doesn't in any was disprove the theorem. You need read more about the subset topology. $\endgroup$– peek-a-booOct 3, 2022 at 4:48
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2 Answers
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Here is the problem: when you write $f:(0,1) \to \{0\}$, you're assuming that the domain $(0,1)$ carries a topology (in this case, it will be the subspace topology induced by the Euclidean topology in the real line). So the conclusion here is that $f^{-1}(\{0\})=(0,1)$ is closed in $(0,1)$. And every subset of a topological space is closed in its own subspace topology.
If the word "topology" doesn't make sense to you yet, the correct conclusion is that $f^{-1}(\{0\})$ is just the intersection of $(0,1)$ with a closed subset of $\mathbb{R}$.
answered Oct 3, 2022 at 4:49
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The precise version of this theorem (in the sense of metric spaces) says the following:
Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces and let $f:X\to Y$ be a function. Then $f$ is continuous if and only if for any set $S$ closed in $Y$, the preimage $f^{-1}(S)$ is closed in $X$.
Note that in this case your space $X$ is the space $(0,1)$ instead of the usual space $\mathbb R$, and $(0,1)$ is always closed in $(0,1)$.
