Construction of a dense subset I have a question. I thought I knew the answer, but for some reasons I can't see how to prove it anymore. Suppose you have a countable dense subset D of $\mathbb R^2$. I want to construct another dense subset of $\mathbb R^2$ with certain properties. From what I remember, if I make sure to choose $b_n\in\mathbb R^2$ such that $d(a_n, b_n) < 1/n$, then the set $\{b_1, b_2, \ldots,\}$ will be dense in $\mathbb R^2$. But I do not remember how to prove this anymore.
Thank you for your help. 
 A: Yes, a countable subset constructed like this will also be dense.
Also, this works for an arbitrary metric space $X$ without isolated points instead of $\mathbb R^2$.
Consider $x\in X$ and $\epsilon>0$. Then there are infinitely many $k\in\mathbb N$ with $d(x,a_k)<\frac\epsilon2$. This is true even if $x\in\{a_k\mid k\in\mathbb N\}$ (why? Note: $x$ is not an isolated point). And among these infinitely many $k$ there are some $>\frac2\epsilon$. Then $d(x,b_k)\le d(x,a_k)+d(a_k,b_k)<\frac\epsilon2+\frac1k<\frac\epsilon2+\frac\epsilon2=\epsilon$.

As the problematic situation with isolated points in the generalization may be somewhat tricky, here's a propositionette that can be used:
Proposition: Let $X$ be a $T_1$  topological space, $A\subseteq X$ a dense subset, $a\in A$ not an isolated point of $X$. Then $A\setminus\{a\}$ is also dense in $X$.
Proof: Let $U\subset X$ be open and nonempty. Since $a$ is not isolated, $U\ne \{a\}$. Since $\{a\}$ is closed, the nonempty set $U\setminus\{a\}$ is open and hence intersects $A$. It follows that $U$ intersects $A\setminus\{a\}$$_\blacksquare$
A: Let $D=\{a_n:n\in\Bbb Z^+\}$ be the original dense set, and let $B=\{b_n:n\in\Bbb Z^+\}$ be such that $d(a_n,b_n)<\frac1n$ for each $n\in\Bbb Z^+$. $B$ is dense in the $\Bbb R^2$ iff each non-empty open set in $\Bbb R^2$ contains a point of $B$.
Let $U$ be any non-empty open set in $\Bbb R^2$, and let $p\in U$. There is an $n\in\Bbb Z^+$ such that $B\left(p,\frac1n\right)\subseteq U$. Since $D$ is dense in $\Bbb R^2$, there is an $m\in\Bbb Z^+$ such that $a_m\in B\left(p,\frac1{2n}\right)$ and $m\ge 2n$. But then $$d(p,b_m)\le d(p,a_m)+d(a_m,b_m)<\frac1{2n}+\frac1{2n}=\frac1n\;,$$ so $b_n\in B\left(p,\frac1n\right)\subseteq U$. Thus, $B$ is dense in $\Bbb R^2$.
This construction works in any separable metric space without isolated points. In general any dense set must include all isolated points.
