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Let $f : X \to Y$ is continuous and $E \subset X$ is dense set. Then I want to show that $f(E)$ is also dense in $f(X)$

My approach is like this :

Let $y \in f(X)\setminus f(E)$, then $\exists x \in X\setminus E $ such that $f(x) = y$.

Then I can find a sequence {${x_n}$} with $x_i \in E$ for all $i$, and $\forall \epsilon > 0, \exists N $ s.t $|x_n -x| < \epsilon$ for $n > N$

I want to conclude that, the sequence $f(x_n)$ converges to $f(x)$. If that is true, since $f(x_i) \in f(E)$ , I can conclude that $y=f(x)\in E'$ which means, $f(E)$ is dense in $f(X)$

However, I'm not sure that conclusion is right. I'm trying to use the definition of continuity but it's confusing to me and feels like I'm wrong. Can you give me an advice?

smw1991
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1 Answers1

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Since $\overline E=X$, for any $x\in X$ there exists a sequence $\{x_n\}$ in $E$ with $\lim_{n\to\infty}x_n=x$. $f$ is continuous, so $f^{-1}(\overline{ f(E)})$ is closed, and as $E\subset f^{-1}(f(E))\subset f^{-1}(\overline{ f(E)})$, we have $x\in f^{-1}(\overline{f(E)})$. It follows that $f(x)\in \overline{f(E)}$, and hence $\overline{f(E)}=f(X)$.

Math1000
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    Thank you. One more question. From the continuity and the sequence {$x_n$} in the question, can we conclude that sequence {${f(x_n)}$} also converges to f(x) in $f(E)$? – smw1991 Mar 30 '16 at 07:54
  • I think that would be a more intuitive way to prove the statement than what I wrote, actually. – Math1000 Mar 30 '16 at 08:02
  • What bothers me is this : continuity is defined like this : $\forall \epsilon$ >0, $|f(x)-f(x_n)| < \epsilon$ implies $\exists \delta$ >0 s.t $|x-x_n| < \delta$. However, what I have in this problem is "$|x-x_n| < \delta$ for any $\delta >0$" part under continuity. can I conclude that $|f(x)-f(x_n)| < \epsilon$? I'm not sure. – smw1991 Mar 30 '16 at 08:12
  • An equivalent definition is that $f:X\to Y$ is continuous iff for each open $U\subset Y$, $f^{-1}(U)$ is open in $X$ (and similarly we may replace "open" with "closed"). – Math1000 Mar 30 '16 at 08:13
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    Thank you very much. I constructed a sequence of neighborhoods of $y \in f(X) \setminus f(E)$ and from the definition you cited, got the result. – smw1991 Mar 30 '16 at 08:25