Show that a subspace of a totally disconnected space is also totally disconnected.

Question

The question is:

Show that a subspace of a totally disconnected space is also totally disconnected.

Definition:

A topological space $X$ is totally disconnected if for any two distinct points $x,y \in X,$ there is a separation $X = U \cup V$ of $X$ with $x \in U $ and $y \in V.$

Definition:

A separation of a space is a cts. function $f: X \rightarrow \{0,1\}$ with $\{0,1\}$ has the discrete topology.

In terms of open sets, a separation of $X$ is an expression $X = U \cup V$ where $U \cap V = \emptyset $ and $U,V$ are both open in $X.$

My questions are:

1- How to prove this statement ?

2- What is the difference between a Hausdoff space and a totally disconnected space?

Could anyone help me answer these questions please?

Answer 1

Let $Y \subseteq X$ and let $y_1, y_2$ be distinct points of $Y$. As these are distinct points of $X$, $X=U \cup V$ with $y_1 \in U, y_2 \in V$ and $U,V$ open and disjoint. But then $U'=U \cap Y$ and $V' = V \cap Y$ are open in $Y$ with the same properties.

For Hausdorff spaces we only know $U \cap V = \emptyset$ but nothing about their union. There are many connected Hausdorff spaces, like $\Bbb R$ and $[0,1]$ etc. But totally disconnected implies both disconnected and Hausdorff. It's a heavier condition. Examples include zero-dimensional spaces like $\Bbb Q$ and the irrationals, and the Cantor set.

The usual definition of $X$ being totally disconnected is that all subspaces of $X$ of size $>1$ are disconnected. This seems like a weaker condition than the one you use. It's clearly implied by it, though.

Answer 2

Suppose that $X$ totally disconnected and that $Y\subset X$. Take $x,y\in Y$ (with $x\neq y$). Since $X$ is totally disconnected, there is a seperation of $X=U\cup V$, with $x\in U$ and $y\in V$. But then $Y=(U\cap Y)\cup(V\cap Y)$ is a separation of $Y$, $x\in U\cap Y$, and $y\in V\cap Y$.

Note that Hausdorff means that for any two distinct elements $x$ and $y$ of $X$, there are open disjoint subsets $U$ and $V$ of $X$ such that $x\in U$ and $y\in V$. Nothing is said about $U\cup V$. Being totally disconnected means that you can write $X$ as the union of two open subsets $U$ and $V$ of $X$ with $x\in U$ and $y\in V$. It's a much stronger condition.