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Suppose $X$ and $Y$ are topological spaces and $f:X→Y$ is a map. Then $f$ is continuous if and only if $f(\bar A)⊆\overline{f(A)}$, where $\bar A$ denotes the closure of an arbitrary set $A$.

This is a very known theorem, I just could not find any example of topological spaces $X$ and $Y$, a function $f$ and set $A$ and an element $x$ such that $x$ is in $\overline{f(A)}$ and not in $f(\bar A)$, someone know an example?

I found out that this element may be an element on $Y$ on the derived set of $f(A)$ but not in $f(D(A))$ where $D(A)$ is the derived set of $A$. But anything ore than this.

Nomenclature: derived set is the set of the limit points, and limit point is a point such that each neighbourhood (open set containing the element) intersects the set in question.

Martin Sleziak
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Yassin Rany
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6 Answers6

12

Consider $f:(-1,1)\to\mathbb{R}$ with $f(x)=x^2$. Both the spaces $X=(-1,1)$ and $Y=\mathbb{R}$ are endowed with the standard topology. Let $A=(-1,1)$. Then $$ f(\overline{A})=[0,1). $$ But $$ \overline{f(A)}=[0,1]. $$

Note that $\overline{A}$ is the closure of $A$ in $X$.

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Here is a not so simple example. There are probably easier ones, but this is just the first one that popped into mind. Consider $f:\mathbb{R}\to[-1,1]\times[-1,1]$ given by $f(t)=(\cos(t),\sin(\pi t))$. Then you can show that the image of $f$ is dense in the square, and hence $\overline{f(A)}$ is the whole square. However, $f(\overline{A})$ is not the entire square since the map is not surjective. Namely, the intersection of the image of $f$ with any horizontal line segment is countable, hence it is not surjective.

TomGrubb
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As an alternative to Jack's excellent and clear answer, here's a 'smallest' example:

Consider the set $Z:=\{0,1\}$ with the two topologies $$\mathcal{T}_1:=\{\varnothing,\{0\},\{1\},\{0,1\}\}\qquad\text{ and }\qquad\mathcal{T}_2:=\{\varnothing,\{0\},\{0,1\}\}.$$ This gives two topological spaces $X:=(Z,\mathcal{T}_1)$ and $Y:=(Z,\mathcal{T}_2)$, and then the identity map $$f:\ X\ \longrightarrow\ Y:\ x\ \longmapsto x,$$ is an example of what you're looking for; take $A:=\{0\}$, then the closure of $A$ in $X$ equals $\{0\}$, whereas the closure of $A$ in $Y$ equals $\{0,1\}$, so $$\{0\}=f(\overline{A})\supsetneq\overline{f(A)}=\{0,1\}.$$

Servaes
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  • really good example thank you, maybe just to not confuse others i think the inclusion sign is directed in the other sense than the wanted direction. Anyway thank you. – Yassin Rany Dec 18 '16 at 12:02
  • Thank you, and thank you for the correction. – Servaes Dec 18 '16 at 13:37
  • There's no need for ${0}$ to be open in the codomain. If you discard it from the topology, you get an example of my answer. And after that you can also discard ${0}$ as an open subset of the domain – Emilio Dec 18 '16 at 14:31
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Let $Y=\mathbb R$ and $A=X=(0,1).$ Define a continuous function $f:X\to Y$ by setting $f(t)=t$ for all $t\in(0,1).$ Then $\overline{f(A)}=[0,1]$ but $f(\overline A)=f(A)=A=(0,1),$ so $0\in\overline{f(A)}$ but $0\notin f(\overline A).$

bof
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For the sake of simplicity, $X$ and $Y$ can be any sets and $f$ can be any function, and (if $Y$ has at least two elements) setting the discrete topology in $X$ and the trivial topology in $Y$ will do the trick: the function will be continuous and for any $x\in X$, $f(\overline { \{x\}})=\{f(x)\} $ will be a proper subset of $\overline{ f(\{x\})}=Y$.

(This is inspired in @Servaes 's answer)

Emilio
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(1). Let $X$ be a dense proper subspace of $Y.$ Let $f=id_X$ and $A=X.$

For example let $X=\mathbb Q$ and $Y=\mathbb R$.

(2). Continuous functions are not generally closed mappings. They need not map closed sets onto closed sets. Any non-closed continuous mapping will provide an example. The projection of $\mathbb R^2$ to its first co-ordinate is continuous. The set $\{(x,1/x):x\ne 0\}$ is closed in $\mathbb R^2.$ Its image under the projection is $\mathbb R$ \ $\{0\},$ which is not closed in $\mathbb R.$

DanielWainfleet
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