Proof. Let \(X\) be separable, then \(X\) has a countable dense subset \(E\). Let \(x\in X\) and \(\varepsilon >0\) be given. Since \(E\) is dense in \(X\), we have that \(\overline {E}=X\) and \(x\in \overline {E}\) so that the open ball \(B_{\varepsilon }(x)\) contains a point \(y\in E\). Therefore \(d(x,y)<\varepsilon \).

Conversely, if \(X\) has a countable subset \(E\) with the given property that every \(x\in X\) is a point of \(E\) or limit point of \(E\). Hence \(\overline {E}=X\).

Therefore \(X\) is separable.

A subset \(A\) of a metric space \(X\) is said to be dense in \(X\) if \(\overline {A}=X\). This is equivalent to the following:

i

For each \(x\in X\) and \(\varepsilon >0\), there exists an element \(a\in A\ni d(x,a)<\varepsilon \).

ii

For \(x\in X\), there exist \(\{a_n\}\in A\) such that \(a_n\rightarrow x\).(Every point \(x\in X\) is a limit point of \(A\)).

iii

Any open set in \(X\) contains a point of \(A\).

Solution. Let \(E\) be the subset of \(C_0\) consisting of all sequences of rational numbers in which only finitely many terms are non-zero. This set is a countable subset of \(C_0\).

To prove that \(E\) is dense in \(C_0\), let \(x\) be any point of \(C_0\); we must find a point of \(E\) arbitrarily close to it. That means choosing \(r=(r_1,r_2,\dots ,r_k,0,0,0,\dots )\) with each \(r_i\) rational and \(\sup _k|x_k-r_k|\) small.
Let \(\varepsilon >0\) be given, we must find an \(n_0\) such that \(|x_n|<\varepsilon \) \(\forall n\geq n_0\).
Let \(r_1,r_2,\dots ,r_{n_0}\) be rational numbers such that \(|x_n-r_n|<\varepsilon \) \(\forall n=1,2,3,\dots ,n_0\). Then \((r_1,r_2,\dots ,r_{n_0},0,0,0,\dots )\) is the required point.

Solution. Let \(A\) be the set of sequences all of whose terms are \(0\) or \(1\). Every such sequence is bounded, so \(A\subset l_{\infty }\).

\(A\) is uncountable

A member of \(A\) is exactly a choice of \(0\) or \(1\) at each of countably many places, so \(A\) is in one-one correspondence with the collection of all subsets of \(\mathbb {N}\), which Cantor’s diagonal argument shows to be uncountable.

Its points are far apart

If \(x,y\in A\) are distinct they differ in some coordinate, and there the difference is \(1\), so \[d(x,y)=\sup _i|x_i-y_i|=1 .\] Hence the balls \(B_{1/3}(x)\), for \(x\in A\), are pairwise disjoint: a point lying in two of them would put two members of \(A\) within \(\frac {2}{3}\) of each other.

No countable set can be dense

Let \(E\) be dense in \(l_{\infty }\). Each ball \(B_{1/3}(x)\) is non-empty and open, so it must contain a point of \(E\), and since the balls are disjoint these points are distinct. That assigns to each of the uncountably many \(x\in A\) its own point of \(E\), so \(E\) is uncountable.

So \(l_{\infty }\) has no countable dense subset. It is worth contrasting this with \(C_0\) above: there the approximating sequences were allowed to be eventually zero, and a \(0\)–\(1\) sequence is never approximated by one of those.

Proof. We prove this in three steps

Step 1

The set of all polynomials is dense in \(C[a,b]\). This is a consequence of Weierstrass approximation theorem which states that for any continuous function \(f(t)\) on \([a,b]\), given \(\varepsilon >0\) there exists a polynomial \(p(t)\) such that \(\displaystyle {\max _{a\leq t\leq b}|f(t)-p(t)|<\varepsilon }.\)

Thus for any \(f\in C[a,b]\) and \(\varepsilon >0\) we can find a polynomial \(p\in C[a,b]\) such that \(d(f,p)<\dfrac {\varepsilon }{2}\).

Step 2

Given a polynomial \(p(t)\), there is a polynomial \(r(t)\) with rational coefficients such that \(\displaystyle {d(p,r)=\max _{a\leq t\leq b}|p(t)-r(t)|<\frac {\varepsilon }{2}}.\hspace {0.2cm}\) Hence \[d(f,r)\leq d(f,p)+d(p,r)<\frac {\varepsilon }{2}+\frac {\varepsilon }{2}=\varepsilon .\] Thus the set of rational polynomials is dense in \(C[a,b]\).

Step 3

The set of all polynomials with rational coefficients is countable thus \(C[a,b]\) is separable. □