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One definition of continuity is the condition $$f(\overline{A})\subseteq \overline{f(A)},$$ for all $A\subseteq X$. To understand this condition better, I tried to find an example of a real-valued function $f\colon\mathbb{R}\to\mathbb{R}$ and a set $A\subseteq\mathbb{R}$ where this inclusion is strict, but I couldn't find one.

If we let $\mathbb{R}_d$ be the real line with the discrete topology, then the identity map $I\colon\mathbb{R}_d\to\mathbb{R}$ and $A = (0,1)$ is an example with strict inclusion. But this example uses a nonstandard topology on $\mathbb{R}$.

  • My question: Is there an example of a continuous function $f\colon\mathbb{R}\to\mathbb{R}$ and set $A\subseteq\mathbb{R}$ with the usual topology on $\mathbb{R}$ such that $$f(\overline{A})\neq \overline{f(A)}?$$
Martin Sleziak
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mars34
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  • You can map the $x$-axis onto a circle minus a point. – Michael Burr Jul 13 '15 at 01:11
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    Observe that if $\overline{A}\subseteq\Bbb{R}$ is bounded, then it is compact. And if $f$ is continuous, then $f(\overline{A})$ is also compact, hence closed. Therefore $\overline{f(A)}\subseteq f(\overline{A})\subseteq \overline{f(A)}$, and we have equality. So any example with the usual topology of $\Bbb{R}$ must use an unbounded $A$. – Jyrki Lahtonen Jul 13 '15 at 12:27
  • @Jyrki Thanks for this observation. It's very interesting. I didn't expect that bounded $A\subseteq\mathbb{R}$ would always fail. – mars34 Jul 13 '15 at 17:28

2 Answers2

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$f(x)=\frac{1}{1+\exp(x)}$, $A=\mathbb{R}$, then $\bar{A}=A$ and $f(A)=f(\bar{A})=(0,1)\neq[0,1]=\overline{f(A)}$.

Kim Jong Un
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Yes. $\;\;\;$ $\hspace{.04 in}f(x) = \dfrac1{1\hspace{-0.05 in}+\hspace{-0.04 in}\left(\hspace{-0.02 in}x^2\hspace{-0.04 in}\right)} \:$ and $\: A = \mathbb{R}$