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I am given a product topology of uncountable $X_a$ (R as index) each homeomorphic to [0, 1]. The product toplogical space is compact and normal but not a metric space. So the question is, is there a subset of the product topological space that is not closed but sequentially closed? I do not think so, but just want to make sure.

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  • $\begingroup$ Hint: consider sequences of ones and zeros. $\endgroup$
    – user87690
    May 7 '20 at 15:05
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There are such sets. A good hint why such sets exist is that despite $[0,1]$ being first countable, uncountable products usually do not preserve first countability, and non-first countable spaces have a good chance of being non-sequential - but that is not sufficient.

A concrete example is the set $A:= \{(x_i)_{i \in \mathbb{R}} \vert x_i = 1 ~ \text{for countably many} ~ i, x_i = 0 ~ \text{otherwise}\}$. Then $A$ is sequentially closed, but not closed. To show this, we show that an arbitrary convergent sequence in $A$ has its limit point in $A$, but there are convergent nets, whose limit point does not lie in $A$.

Then let $(x^n)_{n \in \mathbb{N}} \subseteq A$ be an arbitrary convergent sequence with limit $y \in [0,1]^{\mathbb{R}}$, then the set of $i$s s.t. $y_i = 1$ is countable. Recall that $y_i = 1$ is by definition equivalent to $x_i^n = 1$ for large enough $n$. Hence

$$ \{i \in \mathbb{R} : y_i = 1\} \subseteq \{i \in \mathbb{R} : x_i^n = 1, n \geq m, m \in \mathbb{N}\} = \cup_{m \in \mathbb{N}} \cap_{n \geq m}^{\infty} \{i : x_i^n = 1\}$$

where the $\{i : x_i^n = 1\}$ is countable for each $n \in \mathbb{N}$ by assumption.

However, consider the net $(x^j)_{j \in \mathbb{R}}$ indexed by $\mathbb{R}$ defined by $x^j_i = 1$ when $i \leq j$ and $0$ otherwise, then this net converges to constant $1$ which is not in $A$. Hence this net has a limit which is not in $A$ and thus $A$ is not closed.

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  • $\begingroup$ umm so A is not closed since (1,1,1,1,1,1,1,1,1,1....) is an accumulation point? $\endgroup$
    – ph8ndstne
    May 8 '20 at 0:25
  • $\begingroup$ also I'm having hard time understanding why y has to have countable 1 coordinates $\endgroup$
    – ph8ndstne
    May 8 '20 at 1:40
  • $\begingroup$ I have edited the answer to adress you comments $\endgroup$ May 8 '20 at 7:08
  • $\begingroup$ thank you very much! $\endgroup$
    – ph8ndstne
    May 8 '20 at 12:07

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