Prove directly that the space $\mathbb{R}$ of real number with usual absolute value metric is second countable.

Question

I can do this by proving $B=\{N_q (x):$ $x$ and $q$ are rational $\}$ form a countable basis.

Since $x,q \in \mathbb Q$, and rational is countable so the above set is countable.

Since the union of members of $B$ is $X$ and intersection of members of $B$ is also a member of $B$ since $B$ is finite and finite intersection of open sets is open.

Let $N_1,N_2∈B$, then $N_1=N_{1/a} (x)$ and $N_2=N_{1/b} (x)$ . Suppose $a≤b$, then $1/a≤1/b$, so $N_{1/a} (x)⊆N_{1/b} (x)$, thus $N_1∩N_2=N_1∈B$

So $B$ is a basis.

I'm not sure if this argument solid enough or making sense?

Then the only correction I would suggest is to show that for each $z \in N_q(x)\cap N_r(y)$ there is an element of your base containing $z$ and contained in $N_q(x)\cap N_r(y)$. Your argument along these lines is incomplete and confusing as it is written. ($B$ is not finite, right?) — wckronholm, Mar 20 '14 at 17:41
ok, so you want me to show that $B$ is a open neighbor hood system? — Diane Vanderwaif, Mar 20 '14 at 17:43
Not exactly. There are two conditions to show a collection of open sets $B$ is a base for a topology. You showed that your sets $B$ cover $\mathbb{R}$. The second condition is the one I described above. — wckronholm, Mar 20 '14 at 17:46
Alternatively, you could explicitly show that if $B_1, B_2 \in B$, then $B_1 \cap B_2 \in B$. — wckronholm, Mar 20 '14 at 17:50
I updated my proof, it doesn't look neat, but is it more solid now? — Diane Vanderwaif, Mar 20 '14 at 17:59
Not quite. Notice that your sets $N_q(x)$ can also be written in interval notation as $(x-q, x+q)$. Take two of these and intersect them. Can you find the center and radius? — wckronholm, Mar 20 '14 at 18:07
ok, let $p,q$ are rational, $p>q$ so $x-q> x-P$ and $x+q <x+p$ so $(x-q,x+q) \cap (x-p,x+p)= (x-q,x+q)$, so the cnter is still x and radius still q. — Diane Vanderwaif, Mar 20 '14 at 18:14

score 1 · Answer 1 · answered Mar 20 '14 at 17:38

1

To finish your argument it suffices to express any ball $N_r (x)$ as a union of balls from $B$ (for arbitrary $x,r \in \mathbb{R}$ with $r>0$. To do this just write

$$ N_r (x) = \bigcup_{q \in N_r (x) \cap \mathbb{Q}} N_{\alpha(q)}(q) $$

where $\alpha(q)$ is any rational number with the property that $N_{\alpha(q)}(q) \subset N_r (x)$.

answered Mar 20 '14 at 17:38

Frank

3,844

Sorry for being late but... one can even prove this statement without the requirement of balls (i.e without a metric), right? Just a topology of $\mathbb{R}$ and the fact that for every $V\subset \mathbb{R}$ is open. – user2820579 Nov 05 '20 at 18:53

score 1 · Accepted Answer · answered Mar 20 '14 at 17:49

Your argument so far is fine, but you are not quite done. Since every element of $\mathcal B$ is open in the usual topology on the real line, then to show that $B$ is a basis for said topology, you should show that (1) for any open $U\subseteq\Bbb R$ and any $y\in U,$ there is some $V\in\mathcal B$ such that $x\in V\subseteq U,$ and (2) for any $V_1,V_2\in\mathcal B$ and any $z\in V_1\cap V_2,$ there is some $W\in\mathcal B$ such that $z\in W\subseteq V_1\cap V_2.$ For the latter, it will suffice (though it isn't necessary) to prove that for any $V_1,V_2\in\mathcal B,$ either $V_1\cap V_2\in\mathcal B$ or $V_1\cap V_2=\emptyset.$ (Do you see why?)

score 0 · Answer 3 · answered Mar 20 '14 at 17:37

Consider the basis for $\mathbb{R}$ as $\mathcal{B}=\{(a,b)\ |\ a,b\in \mathbb{Q}\}$. Show that the usual topology on $\mathbb R$ is same as the topology generated by $\mathcal{B}$. Since $\mathcal{B}$ is countable so $\mathbb{R}$ is second countable.

Prove directly that the space $\mathbb{R}$ of real number with usual absolute value metric is second countable.

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