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I noticed that most proofs given in this answer are from first principle.

However, I am wondering if we can construct an alternative proof, by the following fact:

Theorem: Let $X$ and $Y$ be topological spaces and let $f: X \to Y$ be continuous. If $X$ is connected, then $f(X)$ is connected.

(i.e. the continuous image of a connected set is connected.)

For a topological space $(X, \tau)$ and $E, Cl(E) \in X$ with subspace topology $(E, \tau_E)$ and $(Cl(E), \tau_{Cl(E)})$

If we can find a surjective and continuous map, $f: E \twoheadrightarrow Cl(E)$ (i.e. the image, $f(E)$ is precisely $Cl(E)$), this completes the proof by the above theorem.

However I can't think of a way to construct such $f$.

If we restrict $X$ to be a complete metric space $(X, d)$ with $(E, d)$ being a subspace, the completion of $(E, d)$ is precisely $(Cl(E), d)$.

We know that the completion map is an isometry which is continuous, but I am not sure about the case for $X$ being a general topological space.

Many thanks in advance!

EDIT:

I shall emphasise that my main question is about how to construct such surjective and continuous $f$ from a set to its closure in general.

Newton
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1 Answers1

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Consider the set $X$ with $|X|\ge 2$ and $p\in X$ and define

$$\tau_p=\{U\subset X: U=\emptyset \text{ or } p\in U\}$$

Now consider $E=\{p\}$, then $\textrm{cl}(E) =X$

It's easy to show that $\nexists f:E\to \textrm{cl}(E)$ continuous onto map.

Sourav Ghosh
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  • For $E = {p}$, could you elaborate on why $cl(E) = X$ instead of $cl(E) = E = {p}$ please? I almost got it. For instance, I can understand the following example: if we take the 2-point space $X = {p, 2}$ with the discrete topology, then $\tau_p = {\emptyset, {p}, {p, 2}}$ (i.e. the Sierpinski topology), by definition, $cl(E)$ is the intersection of all $\tau-$ closed sets that contains $E$ (which are $\emptyset$ and ${p, 2}$), which indeed yields the whole space. However I can't quite see how to prove this result for a general $X$. – Newton Mar 16 '23 at 00:34