Sequentially closed but no closed subset? 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.
 A: 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. 
A: It is sufficient to take $X_{\alpha}$ to be the two point set $\{0, 1\}$ with discrete topology, with $\alpha$ in an indexed set $J$ that is uncountable. The product of $X_{\alpha}$ with alpha in $J$ is then not first countable. Every point in this product space can be identify as a function $f:J \rightarrow \{0,1\}$.
For the set $A$ that is sequentially closed but not closed, take
$$A=\{f:J\rightarrow \{0,1\} | f^{-1}(\{0\}) \;\text{is countable}\}.$$
Using the fact that countable union of countable sets is countable, one can show that $A$ is sequentially closed.
To show that $A$ is not closed, prove that the point $f_0$, where $f_0(x)=0$ for all $x$ in $J$, is in the closure of $A$ but is not the limit of a sequence in $A$.
The idea is the same as given in the reply above by Chiusole.
