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Say an infinite set is countably compact (if set $E$ is an infinite countably compact set, it contains at least one limit point within itself). Let $x$ be one such limit point of $E$. My textbook says "if this is to be a compact space, every open set containing $x$ has to contain an infinite number of points from $E$".

Is this a necessary condition? Say the base $\mathfrak{B}$ of the topology contains an open set which contains all but a finite number of points from $E$. Let this set be $A$. $A$ may intersect with other sets of $\mathfrak{B}$. However, each of these intersections has to contain all but finite points $\in A$. If these conditions are satisfied, and $E$ is the ony infinite subset of $X$, then $X$ has a finite subcover for every cover, and is hence a compact space!

$x$ may be contained in just two open sets- $X$ and an open set containing a finite number of points from $E$ along with $x$. This makes $x$ a limit point of the infinite set $E$, makes the infinite set $E$ countably compact by giving it a limit point within itself, and also makes $X$ compact if $E$ is the only infinite subset of $X$.

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    $\begingroup$ It's worth noting that "has to contain all but a finite number of points from $E$" is a much stronger condition than "has to contain infinitely many points (of $E$)." The integers, for example, contain infinitely many points (of the real number line), but do not come close to containing all but a finite number of points of the real number line. $\endgroup$ – Cameron Buie Jun 8 '13 at 17:22
  • $\begingroup$ I think it is important to mention whether or not your book defines compact to include Hausdorff condition or not. $\endgroup$ – tomasz Jun 8 '13 at 17:32
  • $\begingroup$ My book's definition of compact- every open cover of a space has a finite subcover. $\endgroup$ – fierydemon Jun 8 '13 at 17:42
  • $\begingroup$ @AyushKhaitan are you working in metric spaces, or some other class? In that case, see my answer. $\endgroup$ – Henno Brandsma Feb 8 '15 at 15:12
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Something may be wrong:

In fact, Assuming $X$ is $T_1$, if $x$ is a limit of $E$, for any open set $U$ of $x$ contain infinite number of points from $E$. So $E$ don't need to be compact.

Example: Let $X=[0,\omega_1)$. It is a countably compact space, since every limit point is in $X$. However it is not compact. Note that for every limit point $x$ of $X$, its every open nbhd contains infinite pints from $X$.


Here is something related your question which you may be interested in:

Let $X$ be a space and $A$ a subset of $X$. A point $x\in X$ is a point of complete accumulation of $A$ if $|U\cap A|=|A|$ for every open nbhd $U$ of $x$.

A space $X$ is compact iff every infinite set in $X$ has a point of complete accumulation.

Proof: see Here

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  • $\begingroup$ You (Paul) may be interested in the following facts: In ZFC, if $X$ is a topological space such that every infinite subset of $X$ has a point of complete accumulation, then $X$ is compact. In ZF, the following are equivalent: (i) Every compact space has the complete accumulation point property. (ii) Every compact Hausdorff space has the complete accumulation point property. (iii) The Axiom of choice holds. From these facts, we see that (in ZF) if every compact space has the complete accumulation point property, then there are no other spaces with this property.... $\endgroup$ – Cameron Buie Jun 9 '13 at 16:52
  • $\begingroup$ As far as I know, it is still an open question whether every space with the complete accumulation point property is compact without the need for Choice. $\endgroup$ – Cameron Buie Jun 9 '13 at 16:54
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    $\begingroup$ But AoC is true so why worry? :D $\endgroup$ – Cheerful Parsnip Dec 1 '13 at 13:45
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There is are different definitions, for a general space $X$:

  • Every infinite subset $E$ of $X$ has a limit point.
  • Every infinite subset $E$ of $X$ has a $\omega$-limit point.

Here $x$ is a limit point of $E$ when every open set $O$ that contains $x$ contains a point from $E$ that is different from $x$.

And $x$ is an $\omega$-limit point of $E$ when every open set $O$ that contains $x$ contains infinitely many points of $E$.

Clearly the second definition implies the first, because if $O$ contains infinitely many points of $E$, one of them is surely different from $x$. In a $T_1$ space (so where finite sets are closed) the first does imply the second, as is easily checked. And e.g. metric spaces are $T_1$.

An example of a space that satisfies the first definition (sometimes called limit point compact) but not the second, is a countable discrete space (say $\mathbb{N}$) times the indiscrete 2 point set $\{0,1\}$ (the so-called doubling of the space) in the product topology. Then if a(n infinite) set $E$ contains, say, $(n,0)$ then $(n,1)$ is a limit point of $E$, as a basic neighbourhood of that point is $\{(n,0), (n,1)\}$ which contains a point of $E$ that is unequal to itself.

The second definition is called countably compact, and is equivalent to "every countable open cover of $X$ has a finite subcover" (see this answer, e.g.). In metric spaces it's equivalent to compactness, but in general it's not.

A compact space always does satisfy the second (and so the first) definition.

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