Isolated Points in Metric Space

Question

Admit the following set theory result : If X is an infinite set and $\mathbb{Q}$ is the set of rational, then there is a bijection $f: X \longrightarrow X \times \mathbb{Q}$. Conclude from this that one can introduce into every infinite set a metric without isolated points.

Thans in advance.

Answer 1

If $(X,d), (Y,e)$ are metric spaces then the metric $f(\,(x,y),\,(x',y')\,)=d(x,x')+e(y,y')$ generates the topology of the product-space $X\times Y$.

If $Y$ has no isolated points then neither does $X\times Y.$ This holds for any spaces $X,Y.$

So let $d$ be "the" discrete metric, i.e. $d(x,x')=1$ if $x\ne x'.$ And let $Y=\Bbb Q$ with $e(y,y')=|y-y'|.$

The metric $f$ on $X\times \Bbb Q$ is equivalent to the metric $d'$ of the Answer from Berci. That is, $f$ and $d'$ generate the same topology.

Answer 2

Define a metric on $X'=X\times\Bbb Q$ by $$d'((x,r),\,(y,s)):=\left\{\matrix{|r-s|&\text{if }x=y\text{ and }|r-s|<1\\ 1&\text{otherwise}}\right.$$ Verify that this is a metric where no point $(x,r)$ is isolated.

Finally, pull back the metric by the bijection $f:X\to X\times\Bbb Q$, i.e. define $d(x,y):=d'(f(x),\,f(y))$.