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Here is the problem:

Let $f : X \to Y$ be a continuous map between topological spaces. Let $E$ be a dense subset of $X$ (that is, $\operatorname{Cl} E = X$), where $\operatorname{Cl}E$ represents the closure of $E$. Prove: $\operatorname{Cl} f(E) = f(X)$

I know continuity of $f$ implies $f(\operatorname{Cl} E) \subseteq \operatorname{Cl} f (E)$, but I have no idea to show the converse.

Antonio Maria Di Mauro
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Allen Cho
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  • the converse inclusion is not true... as $f(X)$ is not closed in general. Anyway, why do you need it? What is your definition of "dense"? – Simone May 07 '14 at 12:53
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    It's not true. Let $X = E = (0,1)$ and let $f\colon (0,1) \to \mathbb{R}$ be the inclusion. Then $\mathrm{cl}(f(X)) = [0,1]$. – jdc May 07 '14 at 12:55

4 Answers4

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$f[X] = f[\operatorname{cl}(E)] \subset \operatorname{cl}(f[E])$. This implies that every point of $f[X]$ is in the closure of $f[E]$, i.e. $f[E]$ is dense in $f[X]$.

Alternatively from the above inclusion $f[X] \subseteq \operatorname{cl}(f[E])$ we see: $$f[X]\supseteq \operatorname{cl}_{f[X]}(f[E]) = \operatorname{cl}(f[E]) \cap f[X] \supseteq f[X] \cap f[X]=f[X]$$ and hence $\operatorname{cl}_{f[X]}(f[E]) = f[X]$ which says $f[E]$ is dense in $f[X]$.

Henno Brandsma
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  • But, how did you show that $\overline{f(E)}=f(X)$? – RFZ Dec 03 '18 at 16:52
  • @ZFR the set contains $f[X]$ as I showed. – Henno Brandsma Dec 03 '18 at 16:57
  • Sorry but I did not understand. We are going to show that $f(E)$ is dense in $f(X)$, i.e. $\overline{f(E)}=f(X)$, right? By continuity we have only $f(X)\subseteq \overline{f(D)}$. How about the converse? – RFZ Dec 03 '18 at 16:59
  • @ZFR A subset $A$ of $f[X]$ is dense in $f[X]$ iff the closure of $A$ in $Y$ contains $f[X]$. This is what I showed. – Henno Brandsma Dec 03 '18 at 17:04
  • What is your definition of dense subset of topological space? – RFZ Dec 03 '18 at 17:07
  • @ZFR the closure is the whole space. But here we are dealing with relative closure and global closure. This might have caused your confusion. – Henno Brandsma Dec 03 '18 at 17:09
  • I guess this is the problem. Could you elaborate your answer, please? Because it is not so clear for beginners in topology – RFZ Dec 03 '18 at 17:10
  • @ZFR like I said in the answer: every point of $f[X]$ is in the closure of $f[E]$. That’s precisely what had to be shown. – Henno Brandsma Dec 03 '18 at 17:20
  • After some thoughts I came up with the following answer: Since $\overline{f(E)}\subseteq f(X)$ then if we consider $f(X)$ as the subspace of $Y$ then closure of $f(E)$ in $f(X)$ is just intersection of closure of $f(E)$ in $Y$ and $X$ and this intersection is $\overline{f(E)}$, right? – RFZ Dec 03 '18 at 17:21
  • @ZFR that’s also a valid argument, indeed. – Henno Brandsma Dec 03 '18 at 17:22
  • But unfortunately I still dont understand your answer. – RFZ Dec 03 '18 at 17:22
  • Every point of $f(X)$ is in the closure of $f(E)$ it means that $f(X)\subseteq \overline{f(E)}$. But it does NOT mean that $f(E)$ is dense in $f(X)$. I guess that it should be clarified – RFZ Dec 03 '18 at 17:24
  • @ZFR $f(X) \subseteq \overline{f(E)}$ from continuity. The closure of $f(E)$ in $f(X)$ equals the closure of $f(E)$ (in $Y$) intersected with $f(X)$ so this includes $f(X) \cap f(X) = f(X)$. – Henno Brandsma Dec 03 '18 at 17:25
  • @ZFR I expanded the last argument I gave in the answer too. – Henno Brandsma Dec 03 '18 at 17:44
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Yes, this is true.

Suppose $f: X \rightarrow Y$ is a continuous map and $A$ is some dense subset of $X$. To see that $f(A)$ is dense in $f(X)$, we can show that the $f(X)$-relative closure of $f(A)$ equals the whole of $f(X)$.

So, let $y$ be any point in $f(X)$ and consider some open neighbourhood $U$ of $y$ in $f(X)$. Spelling this out, $U$ is open in $f(X)$ iff $U = f(X) \cap V$, where $V$ is some open subset of $Y$. Since $f$ is continuous, the pre-image $$f^{-1}(V) = \{ x \in X \ | \ f(x) \in V \}$$ is open in $X$. However, this set is precisely equal to $f^{-1}(U)$. So, we can conclude that $f^{-1}(U)$ is open in $X$.

Since $A$ is dense, it must be the case that $f^{-1}(U) \cap A$ is non-empty (think: dense subsets are "close" to everything). Alternatively, this can be argued formally as follows: since $y \in f(X)$, there is some $x \in X$ such that $f(x) = y$. Thus $x\in f^{-1}(U)$, and then $Cl(A) = X$ implies that $x \in Cl(A)$ and thus $f^{-1}(U) \cap A \neq \emptyset$.

Then any element in the intersection $f^{-1}(U) \cap A$ will get mapped into the set $U \cap f(A)$ under $f$. Hence the set $U \cap f(A)$ is non-empty. Since we chose $U$ arbitrarily, it follows that $y$ lies in the $f(X)$-relative closure of $f(A)$. Then, since we chose $y$ arbitrarily, we may conclude that $f(X) = Cl(f(A))$, that is, $f(A)$ is dense in $f(X)$.

Doc
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If $f(X)$ is closed in $Y$, then $cl(f(E))\subset cl(f(X))=f(X)$ and since $f(X)=f(cl(E))\subset cl(f(E))$, then $cl(f(E))=f(X)$.

user73454
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-8

No. Take constant function on R to R.

Gopi
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