My motivation behind this question is to better under the fundamental concepts of topology in their general sense.

I understand what a set closure when dealing with real numbers and a euclidean metric. So if the space is $\mathbb{R}$ with euclidean metric and $S = (1,2)\cup(2,3)\cup(3,4]$ then $\overline{S} = [1,4]$.But in the more general topological sense i'm not so sure. Next I will show 2 examples where I think I know the answer.

(1) Let $X = ${1,2,3,4,5,6} and $\mathscr{T}_1 = ${{},{1},{1,2},{1,2,3},{1,2,3,4},{1,2,3,4,5},{1,2,3,4,5,6}}. So $(X,\mathscr{T}_1)$ is a topological space.

My claim: If $S = ${1,2,3} then $\overline{S}=X$. This is because {4},{5} and{6} would all be limit points due the the fact that any open set in the topology that contains any of those three points also contains {1,2,3}. Furthermore, for any open set $U \in \mathscr{T}_1$ ;$\overline{U} = X$. Also, in this space, open sets are not closed.

(2) Let $X = ${1,2,3,4,5,6} and $\mathscr{T}_2 = ${{},{1},{2,3},{1,2,3},{4,5,6},{1,4,5,6},{2,3,4,5,6},{1,2,3,4,5,6}}. So $(X,\mathscr{T}_2)$ is a topological space.

My claim: If $S = ${1,2,3} then $\overline{S}=S$. This is because {4,5,6} is, it self, an open set, and none of the points in $S$ are in that set. So{4},{5} and{6} are not limit points. Furthermore, for any open set $U \in \mathscr{T}_2$ ;$\overline{U} = U$. Also, in this space, all open sets are also closed.

Is everything above correct? Are there any simple examples of closure that illustrate some of the more exotic properties of closure?

  • $\begingroup$ Here's a question [math.stackexchange.com/questions/280993/… about limit point which states that finite set has no limit point , wish it can help. $\endgroup$
    – DuFong
    Jun 30 '16 at 18:37
  • $\begingroup$ @Benjamin That question is about the metric space $\Bbb R$. $\endgroup$
    – snulty
    Jun 30 '16 at 18:50

Yes you're approach seems correct. Sometimes I get confused by all the definitions of limit point, accumulation point, adherent point, point of closure, of a set - partly because some require that the point be either in/not in the set, and (open) neighbourhoods of the point have a nonempty intersection with the set, with at least one or two points in the intersection, and so on $\ldots$

Certainly a good definition of closure is the following which is fairly standard. Let $A\subset X$ a topological space with topology $\mathscr{T}_X$,

$$\mathrm{cl}(A)=\bigcap_{\begin{align}C &\text{ closed}\\ &A\subseteq C\end{align}} C$$

In fact as an alternative, in your examples you can compute this by hand. In $\mathscr{T}_1$ the only closed set containing $S$ is $X$ itself so $\mathrm{cl}(S)=X$. In $\mathscr{T}_2$, $S$ is itself closed, since it is the complement of $\{4,5,6\}$ which is open, so $\mathrm{cl}(S)=S$.

$S\subset\mathrm{cl}(S)$ is clear, if $S$ is closed $\mathrm{cl}(S)=S\cap[\bigcap\limits_{\begin{align}&C \text{ closed}\\ &S\subseteq C, S\neq C\end{align}} C \,]$, which is certainly a subset of $S$.

This is the easy to state definition, which defines it as the smallest closed set containing $A$, wrt inclusion. Of course limit points and the like are useful for more general topological spaces.

Some properties of $\mathrm{cl}()$:

$\mathrm{cl}(A\cap B)\subseteq \mathrm{cl}(A)\cap\mathrm{cl}(B)\quad -$ eg. $\mathrm{cl}((-1,0)\cap (0,1))=\emptyset \subset [-1,0]\cap[0,1]$

$\mathrm{cl}(A\cup B)= \mathrm{cl}(A)\cup\mathrm{cl}(B)$

$\mathrm{cl}(\bigcup_\alpha A_\alpha)\supseteq \bigcup_\alpha\mathrm{cl}( A_\alpha) \quad-$ has to do with countable unions of closed set not always being closed.

e.g $\bigcup_{n>1} [0,1-{1\over n}] =[0,1) \subset [0,1]$

You can have that for a disconnected set $D$ that $\mathrm{cl}(D)$ can be connected, like two open disks whose boundary circles intersect at one point only.

You can supplement the idea of closure with the idea of limit points, boundary points etc., to obtain a statement of the form

$$\mathrm{cl}(A)=A\cup\{\text{suitable set}\}$$

There's also the idea of closure with respect to a set say $A\subset B\subset X$, in which case $\mathrm{cl}_B(A)\neq \mathrm{cl}_X(A)$ in general. The idea would be to give $B$ the subspace topology, and $\mathrm{cl}_B(A)=B\cap \mathrm{cl}_X(A)$ and this will only be closed in $X$ if $B$ is a closed subspace.

e.g. $A=(0,1)\subset \Bbb R$ with subspace topology. $\mathrm{cl}_A(A)=(0,1)=A\cap\mathrm{cl}_{\Bbb R}(A)=(0,1)\cap [0,1]$

  • $\begingroup$ Thanks, your example: e.g $\bigcup_{n>1} [0,1-{1\over n}] =[0,1) \subset [0,1]$ was an example I was wondering about. $\endgroup$ Jul 1 '16 at 22:48
  • $\begingroup$ I'm also looking for an example like that but with intersections rather than unions. $\endgroup$ Jul 1 '16 at 22:50
  • $\begingroup$ @MichaelMaliszesky it might be tougher, because arbitrary intersections, countable, uncountable etc are always closed. In the same way arbitrary unions of opens sets are always open. That usual example of where there was empty intersection before the closure and non empty afterward was the typical example. I'll try think of something for tomorrow $\endgroup$
    – snulty
    Jul 2 '16 at 0:28
  • $\begingroup$ Thanks alot, this is a big help. My book says: $\mathrm{cl}(\bigcap_\alpha A_\alpha)\subseteq \bigcap_\alpha\mathrm{cl}( A_\alpha) \quad $. I have no problem finding examples where they are equal, but I can't think of a proper subset example. (I was looking for something like this, when I mentioned "exotic" examples before) $\endgroup$ Jul 2 '16 at 0:36
  • $\begingroup$ @MichaelMaliszesky here's something actually, $\mathrm{cl}(\bigcap_{n>1} (1-\frac{1}{n},1))=\emptyset \subset \{1\}$ $\endgroup$
    – snulty
    Jul 2 '16 at 0:43

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