"Let a,b∈R, with a < b. Prove that each discrete subspace of R and each of the spaces [a,b], (a,b), [a,b),(a,b], (−∞,a), (−∞,a], (a,∞), [a,∞), and {a}, with the topology induced as a subspace of R, is a Polish Space."
This question is a bit confusing. What exactly are discrete subspace of R? Is it just any subset or R with the discrete topology?
Also, Polish space means separable and completely metrizable, but just because R is complete under the usual Euclidean metric doesn't mean a subspace is complete so I must be confusing something.
Best Answer
Each discrete subspace $D$ of $\Bbb R$ is at most countable, and has the discrete topology as its subspace topology (that's the definition of being discrete), and this topology can be induced by the discrete metric $d_1(x,y)=1$ whenever $x \neq y$, and this metric is complete (every $d_1$-Cauchy sequence is eventually constant, and hence converges). So $D$ is Polish (separable, due to its size) and completely metrisable.
Indeed $\Bbb R$ itself is Polish (it's usual metric will show that, and $\Bbb Q$ is countable and dense), and hence so is every closed susbet of $\Bbb R$: we just inherit the metric and the subspace is still sepaarble and compleet in the inherited metric, by standard facts. This handles sets like $[a,b]$, $[a,\infty)$ and $(-\infty,a]$ e.g. and also $\{a\}$ (it's discrete too BTW).
$[a,b) \simeq [a,\infty)$ so if one is Polish so is the other (being Polish only dpends on the topology; it's preserved by homeomorphism), etc. $(a,b) \simeq \Bbb R$ is Polish for the same reason.