# Limit point compact subspace of Hausdorff space

I've encountered this problem in Munkres General Topology, the problem ask if $$X$$ is a limit point compact set, then if $$X$$ is a subspace of a Hausdorff space $$Z$$, is $$X$$ necessarily closed?

I think I know the answer should be wrong, as there is a counter example here: Is a limit point compact subset of a Hausdorff space necessarily closed?

However, I'm not satisfied by it because it make use of the minimal uncountable set, which has no practical example. I've also worked up a proof which proves this implication is true. My idea is to show that every convergent sequence in $$X$$ converges to a point in $$X$$.

Proof: Let $$\{x_1,x_2,\cdots\}$$ be a convergent sequence in $$X$$ with limit $$x\in Z$$, if the sequence is a finite set, then $$x=x_n$$ for all sufficiently large $$n$$, which proves $$x\in X$$. Otherwise if the sequence is an infinite set, from the fact that $$Z$$ is Hausdorff, the sequence must have a unique limit $$x$$. By definition the sequence $$\{x_1,x_2,\cdots\}$$ has a limit point in $$X$$, as stated before this sequence can have only one limit point, so this limit point must be $$x$$, which is in $$X$$. So I've proved $$X$$ is closed because every convergent sequence in $$X$$ converges in $$X$$.

Can you point out which part of my proof is wrong? I really want to believe my proof is wrong, so I can put my effort into learning the minimal uncountable set. It will be better if you can provide a counterexample which doesn't make use of the minimal uncountable set.

Let $$Z=[0,1]^{\Bbb R}$$, all functions from $$\Bbb R$$ to $$[0,1]$$ in the product (aka pointwise) topology which is compact Hausdorff.

Let $$X$$ be its subspace of all functions $$f$$ that have at most countably many non-zero values, i.e. such that $$C(f) = \{x \in \Bbb R\mid f(x) \neq 0\}$$ is at most countable. This $$X$$ is dense in $$Z$$ (so in particular not closed) and limit point compact, even sequentially compact (which is stronger). This is an example which does not require ordinals, and is quite natural IMHO.

• The concept of denseness is a bit strange for me, the book haven't write that far. The compactness of $[0,1]^{\Bbb R}$ is a result of Tychonoff's theorem right? Which I also haven't learn at this stage. But thanks for your answer :) Sep 8, 2020 at 9:31
• @kelvinhong方 I just mean the closure of $X$ is the whole product. So $X$ is not closed; you don’t need that it’s dense per se. You do need to know that a countable product of copies of $[0,1]$ is (sequentially) compact. Sep 8, 2020 at 17:22
• Thanks. It is a clear explanation. Sep 9, 2020 at 10:01

In a general topological space (even a Hausdorff one) closedness of a subset cannot be expressed in terms of sequences. If the limit of every sequence in $$X$$ belongs to $$X$$ you cannot conclude that $$X$$ is closed. Such an argument works in metric spaces but not in general topological spaces.

• I see, so closedness of topological space can only be characterized by limit point, instead of sequence? Sep 4, 2020 at 5:58
• In general topological spaces we have to use nets instead of sequences. But you start with a net in $X$ you cannot use your hypothesis. @kelvinhong方 Sep 4, 2020 at 6:00
• Thanks! Now I should examine the minimal uncountable set. Sep 4, 2020 at 6:01