1

I am aware that this question has been asked multiple times on this site; however none of them seem to have a proof as simple as mine. So I would like to know if this is legitimate:

Let $A=\{x_n:n\in\mathbb Z_+\}$ be a countable dense set in $X$, then $\overline A=X$. So if $x\in X$, then $x\in A$ or $x\in \overline A$. Either way, any neighborhood $U$ of $x$ contains a point of $A$. So there exist positive integers $n,m$ such that $B_d(x_n,1/m)\subset U$, with $x\in B_d(x_n,1/m)$. Hence the set $$B =\{B_d(x_n,1/m):n,m\in\mathbb Z_+\}$$ is a countable basis for $X$.

Math1000
  • 36,983
  • The third sentence doesn't seem to follow from the first two sentences. – Matt Samuel Nov 18 '19 at 21:05
  • 2
    (7)$\to$(1) in this one is more general, clear, simpler and complete, since you haven't shown that the ball $B_d(x_n,1/m)$ satisfies both being in $U$ and containing $x$. Maybe once one chooses $1/m$ to be inside $U$ then $x$ escaped outside of the ball. The $x_n$ must be carefully chosen to ensure that $m$ can be chosen to satisfy both. – conditionalMethod Nov 18 '19 at 21:06
  • @conditionalMethod That is a good point. How can I fix the proof? I am not interested in further generality at this point, only the claim at hand. – Math1000 Nov 18 '19 at 21:10
  • In the link is the argument choosing conveniently the $x_n$ first (it is called $y$ there and $A$ is called $D$). Just replace any mention of cardinality by 'countable'. Well, they worked with rational radii, which gives more freedom to choose. The radii $1/m$ can be made to work too, but one might need to choose smaller fractions of $\epsilon$ than $\epsilon/3$ and $\epsilon/2$. Explicitly, if the $r$ chosen there was $p/m$, then for your collection of balls maybe work with $\epsilon/(3p)$ and $\epsilon/(2p)$. – conditionalMethod Nov 18 '19 at 21:18
  • The $3$ and $2$ are of course, nothing special. Drawing a picture helps, because the idea is to choose the $x_n$ so close to $x$ inside the ball $B(x,\epsilon)\subset U$ that one can draw a ball with center $x_n$ an radius slightly larger than $d(x,x_n)$ and remain inside $B(x,\epsilon)$. Testing this requirement with the triangle inequality. – conditionalMethod Nov 18 '19 at 21:26
  • Got it. I will revise my proof and post it as an answer then. – Math1000 Nov 18 '19 at 21:27
  • @MattSamuel . You could say it follows.... eventually, after some important (missing) intermediate steps. – DanielWainfleet Nov 18 '19 at 21:53
  • Your proof is incomplete, claim "with $x\in B_d(x_n,1/m)$" was not proven. For this, you could invoke the triangle inequalities. Actually, in any proof of the given theorem you must involve the triangle inequality since the theorem would be false if tr.ineq. were not assumed. – Wlod AA Nov 19 '19 at 03:26

1 Answers1

1

Let $d$ be the metric on $X$. Let $A=\{x_n:n\in\mathbb Z_+\}$ be a countable dense set in $X$, then $\overline A=X$. Put $$\mathcal B = \{B_d(x,r):x\in A,r\in\mathbb Q\},$$ then $\mathcal B$ is countable as $A$ and $\mathbb Q$ are countable. Let $U$ be an open set in $X$ and $x\in U$. Because $U$ is open, we may choose $\varepsilon>0$ such that $B_d(x,\varepsilon)\subset U$, and since $A$ is dense there exists $x_n\in A$ with $x_n\in B_d(x,\varepsilon/3)$. Pick $r\in\mathbb Q$ with $\varepsilon/3<r<\varepsilon/2$, then $x\in B_d(x_n,r)$ since $B_d(x_n,r)\subset B_d(x,\varepsilon/3)$. Moreover, $B_d(x_n,r)\subset B_d(x,\varepsilon)$, since if $d(y,x_n)<r$ then $$ d(x_n,x) \leqslant d(y,x_n) + d(x_n,x) < r + r < \varepsilon. $$ It follows that $x \in B_d(x_n,r)\subset B_d(x,\varepsilon) \subseteq U$, so that $\mathcal B$ is a basis for $X$, as desired, as of course $B_d(x_n,r) \in \mathcal{B}$ by definition.

Henno Brandsma
  • 242,131
Math1000
  • 36,983