Explain the statemets related to separable spaces.

Question

• Separable spaces are "simpler" than nonseparable ones-KREYSZIG
• Separable space $X$ is not "too big" in the sense that we can approach each element of space $X$ through a sequence of elements of a countable set-PONNUSAMY

I'm reading the article of Dense subsets and separability from different texts.I got the above statements which I'm not getting.

1.What notions are involved in non-separable spaces so that they are complex than Separable ones?

2.What is the meaning of "too big" in the second statement?

Answer 1

A separable space is a space with a countable dense subset. So for a non-separable space, for any countable subset, its closure will not be the entire space.

For example $l_{\infty}$ is non-separable because the open balls of radius $\frac{1}{2}$ centred at the elements with only 0 and 1 entries form an uncountable collection and any dense set would have to contain an element of each of these balls. This in particular proves that $R$ cannot continuously surject onto $l_{\infty}$. Intuitively this makes $l_{\infty}$ look rather large.

Having said all these, remember that these are just intuitions in a broad sense and things that are meant to sell the idea of non-separable to you as a new and mostly inexperienced reader. After all, $\mathbb{R}$ and $l_{\infty}$ have the same cardinality so $l_{\infty}$ isn't really larger than $\mathbb{R}$ in this sense. An experienced mathematician doesn't really use these intuitions in proofs or reasonings.

Answer 2

Re: 2.

Not all topologies can be described in terms of sequences. The definition of a topology $T$ on a set $X$ is short and not very restrictive. An important class of topological spaces is the class of sequential spaces, in which the closure of a set can be described in terms of convergent sequences. This class includes all metric spaces.

If $X$ is a sequential space and $D$ is countable dense subset of $X,$ let $E$ be the set of all sequences of members of $D.$ The cardinal $|E|$ of $E$ is at most $2^{\aleph_0}=|\Bbb R|.$ Let $F$ be the subset of $E$ that contains only the convergent sequences. For any sequence $f\in F$ let $\psi(f)$ be the limit of the sequence $f.$ Then, since $D$ is dense in $X,$ we have $X=\{\psi (f):f\in F\}$. Therefore $|X|=|\{\psi(f):f\in F\}|\le |F|\le |E|\le |\Bbb R|.$ So $X$ is not "too big" in terms of its cardinality.

On the other hand if the topology $T$ on $X$ is $T=\{X,\emptyset\}$ then any non-empty $D\subset X$ is dense, so $X$ is separable. But $X$ can be any set so $|X|$ can be any cardinal. And if $X$ has at least $2$ members then $X$ is not a sequential space.

BTW. A topological definition of "$(x_n)_{n\in \Bbb N}$ converges to $x$" is: $\{x\}=\cap_{n\in \Bbb N}\overline {S_n},$ where $S_n=\{x_j:j\ge n\}.$