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$X$ is a topology space and $A=\{x_1 , x_2 , ... , x_n , ... \}$ is a subset of $X$ , $y$ is a limit point of $A$ .
Can we show that for each $N$ , $y$ is also a limit point of $A'=\{x_N ,x_{N+1}, ... \}$ ?

It is obvious that the conclusion hold when $X$ is a metric space . But it might not be valid for all topology space since if it is true , then we can use it to prove : If every infinite subset in $X$ has a limit point , then $X$ is compact . (Which is not true for all topology space )
Suppose there is family of open sets $O_1,\ldots,O_n,\ldots$ such that $X \subset \cup O_n$ but there is no finite subcovering. Construct the sequence $A=(x_n)$ such that $x_n \in X \setminus \bigcup_{i=1}^n O_i $ . But our assumption $A$ has an limit point $x$ , we say that $x \in O_N$ for some $N$ . Let $$A'=\{x_N ,x_{N+1}, ... \}$$ then we can see $x$ is not a limit point of $A'$ , since $x \in O_N $ but $O_N$ contains no element of $A'$ .

My question:
Is there any counterexample such that : $X$ is a topology space , every infinite subset of $X$ has a limit point but $X$ is not compact .

J.Guo
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  • The real numbers? – AlkaKadri Jan 28 '19 at 02:52
  • @AlkaKadri I think $N$ is an infinite set of $R$ which has no limit point . – J.Guo Jan 28 '19 at 02:57
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    Ah very nice. Just thought about it some more, and I think you might want to consider spaces where compact and sequentially compact are not equivalent. For example, https://www.andrew.cmu.edu/user/calmost/pdfs/sasms_F04.pdf and https://math.stackexchange.com/questions/152447/compactness-sequentially-compact – AlkaKadri Jan 28 '19 at 03:01
  • Do you know about ordinal numbers? Well-ordered sets? – bof Jan 28 '19 at 03:30

1 Answers1

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That $x$ is a limit point of $A$ implies that it is a limit point of $A\setminus F$, where $F \subseteq A$ is finite, will hold in any $T_1$ space (so when singletons are closed). That answers the first part of your question.

In a $T_1$ space, being a limit point of $A$ (every neighbourhood of $x$ intersects $A\setminus\{x\}$) is equivalent to $x$ being an $\omega$-limit point of $A$ (every neighbourhood of $x$ contains infinitely many points of $A$).

"Every infinite set $A$ has an $\omega$-limit point" is in general spaces equivalent to "every countable open cover of $X$ has a finite subcover", i.e. countable compactness, which for metrisable spaces indeed is equivalent to compactness but in general not: consider $\omega_1$, the first uncountable ordinal in the order topology, or $\{0,1\}^\mathbb{R}\setminus \{\underline{0}\}$, which are countably compact but not compact. The first is also sequentially compact, first countable, hereditarily normal etc., so even for very nice spaces these notions need not be equivalent.

Henno Brandsma
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