when we need the property of dense

Question

In topology space $(\Omega,\mathcal{A})$, a set $D$ is defined as dense set when cl$(D)=\Omega$. As I know from here, The main aim to use dense to try to find a countable subset $D$ which is dense in $\Omega$. For example, the subset of all rational numbers $\mathbb{Q}$ is dense in the set of all the real numbers $\mathbb{R}$.

My question is: how to take advantage of the property of dense? or if $\Omega$ has a (countable) dense subset, what will happens?

Answer 1

A lot of useful things, two of them:

If $X$ is a locally convex topological vector space, the dual $X^*$ endowed with the weak-$*$ topology has the property that each compact subset $K$ is metrizable.

If $X$ is a metric space, and $m$ is a borel regular measure on $X$, then for each $A$ measurable with $m(A) < \infty$ and each $f:A \to S$ is mesurable, where $S$ is a seprable metric space, Lusin's theorem holds: for each $\epsilon >0$ there exists a closed set $C\subset A$ with $m(A-C)<\epsilon $ such that $f$ is continuous on $C$

Answer 2

More examples:

(1). First, a preliminary theorem: Let $Y$ be a $T_2$ (Hausdorff) space and let $X$ be any space. If $f:X\to Y$ and $g:X\to Y$ are continuous then $\{p\in X: f(p)=g(p)\}$ is a closed subset of $X.$ (Not difficult. I think it is easier to show, equivalently, that $\{q\in X:f(q)\ne g(q)\}$ is open in $X$.)

An immediate corollary is that if $Y$ is $T_2$ and the continuous functions $f:X\to Y$ and $g:X\to Y$ agree on a dense subset of $X,$ then $f=g.$

For example if $f:\Bbb R^m\to \Bbb R^n$ and $g:\Bbb R^m\to \Bbb R^n$ are continuous (with $m,n \in \Bbb Z^+$) and if $f, g$ agree on $\Bbb Q^m,$ then $f=g.$

This corollary is also used in proving the Jones Lemma. A particular case (the smallest case) of the Jones Lemma is that if a separable space $X$ has a closed discrete subspace $Z$ of cardinality $2^{\aleph_0}$ then $X$ is not a $T_4$ (normal) space. For example the Niemitzky plane is not $T_4.$

(2). Theorem: If a metric space $X$ is separable then $X$ has a countable base (basis). Corollary: The topology of the Sorgenfrey line $S$ (which is the set $\Bbb R$ with the "upper-limit topology") cannot be generated by a metric. Because it is easily shown that (i) $\Bbb Q$ is a countable dense subset of $S,$ and (ii) $S$ does not have a countable base.

Footnote: Re (2): If $(X,d)$ is a metric space and $D$ is a dense subset of $X$ then $\{B_d(x,q):x\in D\land q\in \Bbb Q^+\}$ is a base for the space.