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I came across this text in Rudin's book where it has been mentioned that a non-empty finite set is closed.

But a closed set is a set which contains all of it's limit points in the set itself but none of the elements of a non-empty finite set can possibly have a limit point because a neighborhood of a limit point has infinite points . So how come a non-empty finite set be a closed set ??

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    $\begingroup$ A singleton is obviously closed (in the context of Rudin), and then you have that a finite set is the union of finitely many closed sets, and thus closed. $\endgroup$
    – Kwame Gonzales
    Oct 25, 2017 at 11:29
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    $\begingroup$ In the world of general topology, your statement is false, as you only specify a finite set but no topology. However, a non-empty finite subset of, say, $\mathbb{R}$ or $\mathbb{R}^n$ (with the standard topology), is always closed. $\endgroup$
    – MartianInvader
    Oct 25, 2017 at 18:22
  • $\begingroup$ @MartianInvader Well I am considering it in terms of real analysis but thanks for mentioning the fact - I will remember it when I will read topology $\endgroup$
    – Aashish Loknath Panigrahi
    Oct 25, 2017 at 18:56
  • $\begingroup$ By the way, in the more general setting, a topological space in which all finite sets are closed is called a T₁ space. Spaces used in analysis are always T₁. By the way, you do not need the word non-empty, of course, since the empty subset is closed as well. $\endgroup$
    – Jeppe Stig Nielsen
    Oct 25, 2017 at 19:38
  • $\begingroup$ It is also easy enough to directly check that the complement is open, which is another way of seeing why this is true. $\endgroup$
    – John Coleman
    Oct 25, 2017 at 22:18

3 Answers 3

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Disregarding possible fine print of definitions the resolution is quite simple:

If there are no limit points, then of course all limit points are trivially contained in the set.

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"$A$ contains all of its limit points" is short for "for all $x$ such that $x$ is a limit point of $A$, we have that $x \in A$".

This statement is voidly true if there are no limit points (as in this case).

Equivalently put, the condition can be written as $A' \subseteq A$ and the empty set is a subset of all sets.

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  • $\begingroup$ Thanks for the good answer, can we say this set is "open" and "close" simultaneously? Because all the points are interior points and the set contains all of them, therefore it is also an open set. $\endgroup$
    – K.K.McDonald
    Mar 31, 2020 at 20:08
  • $\begingroup$ @K.K.McDonald no, in a finite set not all points are interior points, necessarily. In most spaces they won’t be. $\endgroup$
    – Henno Brandsma
    Mar 31, 2020 at 20:13
  • $\begingroup$ Yes you are right, here in this question Euclidean distance is of interest and thus these points are not interior point, however if we considered the discrete distance i.e. $d(x,y)=0$ if $x=y$ and $d(x,y)=1$ otherwise, then these points are interior points? $\endgroup$
    – K.K.McDonald
    Mar 31, 2020 at 20:18
  • $\begingroup$ @K.K.McDonald in the discrete matrix topology all subsets are open and closed. $\endgroup$
    – Henno Brandsma
    Mar 31, 2020 at 20:19
  • $\begingroup$ Thanks again, I understand this problem now. $\endgroup$
    – K.K.McDonald
    Mar 31, 2020 at 20:22
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Alternatively a set $A$ is closed iff $A=\bar A= A \cup \{\text{set of all limit points}\}.$

In your case, $\{\text{set of all limit points}\}=\emptyset.$ Thus $A=\bar A$.

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