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I found this comment in my lecture notes, and it struck me because up until now I simply assumed that continuous functions map closed sets to closed sets.

What are some insightful examples of continuous functions that map closed sets to non-closed sets?

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    $f(x) = \frac1{1+x^2}$ where $x \in \mathbb{R}$2012-05-21
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    $f(x) = \text{arc tg}(x)$ for $x\in \mathbb{R}$.2012-05-21
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    $f:\mathbb{R}^2\longrightarrow\mathbb{R},f(x,y)=x$2012-05-21
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    Let $X$ be $\Bbb R$ with the discrete metric and $Y$ be $\Bbb R$ with its usual metric. Let $f$ be the identity mapping.2012-05-21
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    $f(x) = e^{x}$ on $\mathbb{R}$.2012-05-21
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    Any strictly monotonic function that is bounded on the appropriate side.2012-05-21
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    Despite continuous functions can map closed sets to non-closed sets, they *always* send compact sets to compact sets.2012-05-21

1 Answers 1

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Here’s a small sample of examples.

The $G$ graph of $y=1/x$ is closed in $\Bbb R^2$, and the map $p:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x$ is continuous, but $p[G]=(\leftarrow,0)\cup(0,\to)$, which is not closed in $\Bbb R$.

The map $\Bbb R\to\Bbb R:x\mapsto e^{-x}$ sends the closed subset $[0,\to)$ of $\Bbb R$ to the non-closed subset $(0,1]$. Other functions with horizontal asymptotes provide similar examples.

If $X$ is any non-closed subset of a space $Y$, the inclusion map $i:X\to Y:x\mapsto x$ gives a trivial example, since $X$ is a closed subset of itself.

Another trivial example is obtained by taking any infinite set $X$, letting $\tau_d$ be the discrete topology on $X$, and letting $\tau$ be any other topology on $X$. The identity map from $\langle X,\tau_d\rangle$ to $\langle X,\tau\rangle$ is automatically continuous. However, there is at least one $x_0\in X$ such that $\{x_0\}\notin\tau$ (i.e., $x_0$ isn’t an isolated point of $\langle X, \tau \rangle$); if $A=X\setminus\{x_0\}$, then $A$ is closed in $\langle X,\tau_d\rangle$ (as is every subset of $X$), but $A$ is not closed in $\langle X,\tau\rangle$.

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    Brilliant, Brian.2013-07-06
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    Thank you. Today it seems so obvious. It is strange to think that a year ago it was not.2013-09-11
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    @Anonymous: You're welcome. Strange, perhaps, but it should be gratifying to realize how far one has come.2013-09-12
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    I like how your 1st example shows linear fns don't even necessarily map closed sets to closed sets2015-08-14