Relation between dense subset and cover of a metric space Suppose a subset of a metric space is dense. Does that imply that neighbourhoods of the elements of that dense subset covers the metric space. Is the converse true ? Can someone explain this ?
 A: One has to be careful. The following example may serve to correct your quite reasonable intuition.
Take as your metric space the reals between $0$ and $1$, with the usual metric. The rationals between $0$ and $1$ are dense in this space.
List these rationals as $r_1,r_2,r_3,\dots$. 
Around $r_1$, put an open interval of total length $\frac{1}{100}\cdot\frac{1}{2}$. Around $r_2$, put an open interval of length $\frac{1}{100}\cdot\frac{1}{2^2}$. Around $r_3$ use an interval of length $\frac{1}{100}\cdot\frac{1}{2^3}$. And so on.
The sum of the lengths of these intervals is $\frac{1}{100}$. It is reasonable to conclude (and indeed true) that the intervals only cover a small fraction of the unit interval. By modifying the construction, one can ensure that any particular irrational is left out.
A: No. For each $q\in\Bbb Q$ let $r_q=|q-\sqrt2|$; then $\Bbb Q$ is dense in $\Bbb R$, and $N(q,r_q)$ is an open nbhd of $q$ for each $q\in\Bbb R$, but $\{N(q,r_q):q\in\Bbb Q\}$ doesn’t cover $\Bbb R$: its union is $\Bbb R\setminus\{\sqrt2\}$. By working a little harder I could have arranged for the union to miss uncountably many points of $\Bbb R$.
In the other direction, note that $\{N(k,1):k\in\Bbb Z\}$ is an open cover of $\Bbb R$, but $\Bbb Z$ is very far from being a dense subset of $\Bbb R$: it’s a closed set with empty interior.
(Here $N(x,r)=\{y\in\Bbb R:|x-y|<r\}$, the open ball of radius $r$ centred at $x$.)
