# Sequential spaces, IV

Uncountable product of first countable spaces can never be first countable (see The product of first countable spaces). It turns out that uncountable products of first countable spaces cannot even be sequential. In this post we show that the product of uncountably many Hausdorff spaces, each of which has at least two points, can never be sequential. This follows from the fact that the product space $2^S$ is not sequential where $2=\left\{0,1\right\}$ is the two-point discrete space and $S$ is any uncountable set. The space $2^S$ can be embedded as a closed subspace of a product of uncountably many Hausdorff spaces, each of which has at least two points.

For discussion on this blog about sequential spaces, see Sequential spaces, I, Sequential spaces, II and Sequential spaces, III.

Let $S$ be an uncountable set. Let $X=2^S$ be the product of uncountably many copies of $2=\left\{0,1\right\}$ indexed by the set $S$. Let $Y$ be all points $x \in X$ such that $x(\alpha)=x_\alpha=0$ for all but countably many $\alpha \in S$. In other words, $Y$ is the $\Sigma$-product of $2=\left\{0,1\right\}$. Clearly $Y$ is not closed in $X$. We claim that $Y$ is sequentially closed in $X$.

For each $z \in Y$, let $W_z=\left\{\alpha \in S:z_\alpha \ne 0\right\}$. Note that each $W_z$ is countable. Suppose that $\left\{y_n\right\}$ is a sequence of points of $Y$ such that $y_n \rightarrow y \in X$. We show that $y \in Y$. Let $W$ be the union of all $W_{y_n}$, which is a countable set. For all $\alpha \in S-W$, $y_n(\alpha)=0$ and thus $y(\alpha)=y_\alpha=0$. Thus $y \in Y$. This shows that $Y$ is sequentially closed in $X$.It follows that $2^S$ is not sequential.

As stated at the beginning of the post, any uncountable product of Hausdorff spaces, each of which has at least two points, can never be sequential. Consequently, the property of being a sequential spaces is not preserved in uncountable products.