2
$\begingroup$

Let $X$ be a topological space and $A \subseteq X$ a subset. $A$ is called sequentially compact iff every sequence in $A$ has a convergent subsequence with limit in $A$. A point $x \in X$ is an $\omega$-accumulation point of $A$ iff every open neighborhood of $x$ contains infinitely many points of $A$.

Is the following statement true?

$A$ is sequentially compact $\Leftrightarrow$ every countably infinite subset $B \subseteq A$ has an infinite subset $B' \subseteq B$ that has an $\omega$-accumulation point $x \in X$.

Proof: "$\Rightarrow$": Let $B \subseteq A$ countably infinite. Then there is a bijection $x : \mathbb{N} \to B$ which gives us a sequence $(x_n)$ in $B$ with distinct values $x_n$ such that $B = \{ x_n \mid n \in \mathbb{N} \}$. Since $A$ is sequentially compact there is a convergent subsequence $(x_{n_k})$ with limit $x \in A$. Since all the $x_n$ are distinct the set $B' := \{ x_{n_k} \mid k \in \mathbb{N} \}$ is an infinite subset of $B$ and the limit $x$ is clearly an $\omega$-accumulation point of $B'$.

"$\Leftarrow$": Let $(x_n)$ be a sequence in $A$. If $B := \{ x_n \mid n \in \mathbb{N} \}$ is finite then some value $x \in B$ of $(x_n)$ is taken infinitely often and we can take the constant subsequence $x_{n_k} := x$ that converges to $x \in A$. So assume that $B$ is (countably) infinite. By the premise, the countably infinite set $B$ contains an infinite subset $B' = \{ x_{n_k} \mid k \in \mathbb{N} \}$ that has an $\omega$-accumulation point $x \in X$. Now the sequence $(n_k)_k$ is in general not strictly increasing such that $x_{n_k}$ is not a subsequence of $x_n$. However, $n_k$ has a strictly increasing subsequence $n_{k_l}$ which gives us a subsequence $x_{n_{k_l}}$ of $x_n$. Set $C := \{ x_{n_{k_l}} \mid l \in \mathbb{N} \} \subseteq B'$.

Now here I am stuck. If $x_{n_{k_l}}$ converges, then we are done (if the limit is in $A$). But if $x_{n_{k_l}}$ does not converge, why is then $x$ an $\omega$-accumulation point of $C$? Assuming, $x$ is an $\omega$-accumulation point of $C$ then $x$ is also a cluster point of $x_{n_{k_l}}$ but why does it follow that $x_{n_{k_l}}$ has a further subsequence that converges to $x$? (The last assertion becomes true if $X$ is Fréchet-Urysohn but not in a general topological space).

Remark: I try to generalize the statement in "Handbook of Set-Theoretic Topology" by K. Kunen and J. Vaughan, p. 573 which states in the context of $T_3$ spaces (regular Hausdorff) that sequential compactness is equivalent to "every countably infinite subset has an infinite subset having exactly one (complete) accumulation point). I want to get rid of the $T_3$ assumption which I think ends up in replacing "exactly one accumulation point" by "some accumulation point" (no uniqueness due to loss of Hausdorffness) and "accumulation point" by "$\omega$-accumulation point" due to loss of $T_1$-ness.

$\endgroup$
2
  • 1
    $\begingroup$ Somewhat tangentially re your remark, I don't see how that result needs regularity at all (Hausdorff seems enough to me...). The "exactly one accumulation point" condition is doing much more work than just corresponding to the Hausdorff condition though; it is what guarantees that your subsequence actually converges to the point it accumulates to (since if it didn't converge, you could find a subsequence that doesn't accumulate at the point, and then that subsequence would have to accumulate somewhere else, contradicting uniqueness of the point). $\endgroup$ – Eric Wofsey Feb 4 '16 at 8:44
  • $\begingroup$ @EricWofsey: I see, uniqueness of accumulation points of sets should not be confused with uniqueness of limits of sequences. $\endgroup$ – yada Feb 4 '16 at 9:15
1
$\begingroup$

Every compact space satisfies your second condition: if $B$ is an infinite subset of a compact space $X$, take the intersection of the closures of all the cofinite subsets of $B$. These have the finite intersection property, so their intersection is nonempty, and any point in the intersection is an $\omega$-accumulation point of $B$. So any compact space that is not sequentially compact is a counterexample to your statement.

$\endgroup$
4
  • $\begingroup$ Ok, I see. Every compact space is countably compact. An equivalent condition for countably compactness is "every infinite set has an $\omega$-accumulation point". Thus, for all countably infinite $B$ choose $B' := B$. What I do not understand so far, is where $T_3$-ness is needed in the proof of the equivalence in the referenced book. $\endgroup$ – yada Feb 4 '16 at 9:13
  • $\begingroup$ Well, as I commented, I don't see how $T_3$-ness is needed either either; the proof I came up with only requires Hausdorffness as far as I can see. $\endgroup$ – Eric Wofsey Feb 4 '16 at 9:16
  • $\begingroup$ I'm having some troubles: If $x_n$ is a sequence with distinct values then in a general topological space $x$ is an $\omega$-acc. point of the (infinite) set $\{ x_n \}$ iff $x$ is a cluster point of the sequence $(x_n)$. If $x$ is a (subsequential) limit of $(x_n)$ then $x$ is a cluster point of $(x_n)$ (and the converse does hold in Fréchet-Urysohn spaces, e.g. first-countable spaces). Together, $x$ can be an $\omega$-acc. point of $\{ x_n \}$, i.e. a cluster point of $(x_n)$ but no subsequence converges to $x$. $\endgroup$ – yada Feb 4 '16 at 9:32
  • 1
    $\begingroup$ Right, but you haven't used the full strength of the hypothesis. You know that $x$ is the unique cluster point of $(x_n)$ (having passed to a subsequence). If $(x_n)$ didn't converge to $x$, then there is a neighborhood of $x$ omitting infinitely many of the $x_n$'s, which then form a subsequence which does not cluster at $x$. But the hypothesis (applied to this subsequence) implies it must cluster somewhere, and that cluster point is then also a cluster point of the sequence $(x_n)$, contradicting uniqueness of $x$. $\endgroup$ – Eric Wofsey Feb 4 '16 at 9:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.