Let I be a uncountable set and {$(X_i, \tau_{i} $): i $\in$ I} a family of topological spaces $T_1$ with at least two points

Question

Let I be a uncountable set and {$(X_i, \tau_{i} $): i $\in$ I} a family of topological spaces $T_1$ with at least two points.Set f $ \in \prod_{i \in I} X_i $.Show that the subspace S = { g $\in$ $\prod_{i \in I} X_i$ :( i $\in$ I: g (i) $\neq$ f (i)) is countable} from $ \prod_{i \in I} X_i $ is not separable.

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What I thought.

To be separable $ \prod_ {i \in I} X_i $ needs to be countable and dense, we know that it is countable, so it cannot be dense. We also know that I is $ T_1 $ so there is z $ \in $ I and U $ \subset $ I, where z $ \in $ U such that z $ \notin $ S. So $ \prod_ {i \in I} X_i $ is not dense and therefore is not separable.

Thank you in advance for any assistance.

Answer 1

There are several problems with your attempt. First, we’re not interested in whether $\prod_{i\in I}X_i$ is separable: we’re interested in the separability of its subspace $S$. And in any case separability of $\prod_{i\in I}X_i$ would not require $\prod_{i\in I}X_i$ to be countable, and neither $\prod_{i\in I}X_i$ nor $S$ is countable. Here is a correct argument:

Let $D=\{x_n:n\in\Bbb N\}$ be a countable subset of $S$. For each $n\in\Bbb N$ let $I_n=\{i\in I:x_n(i)\ne f(i)\}$; $x_n\in S$, so $I_n$ is countable. Let $J=\bigcup_{n\in\Bbb N}I_n$; $J$ is the union of countably many countable sets, so $J$ is countable, and therefore $I\setminus J$ is uncountable. In particular, $I\setminus J\ne\varnothing$, so we may pick an $i_0\in I\setminus J$. Let $u\in X_{i_0}\setminus\{f(i_0)\}$, and define $y\in S$ as follows:

$$y(i)=\begin{cases} u,&\text{if }i=i_0\\ f(i),&\text{if }i\ne i_0\,. \end{cases}$$

For $i\in I\setminus\{i_0\}$ let $U_i=X_i$, and let $U_{i_0}=X_{i_0}\setminus\{f(i_0)\}$; then $S\cap\prod_{i\in I}U_i$ is an open nbhd of $y$ in $S$ that is disjoint from $D$, so $D$ is not dense in $S$. $D$ was an arbitrary countable subset of $S$, so $S$ is not separable.