# $(a,b)$ is polish space with induced topology

From Topology Without tears:

Prove that each discrete space and each of the spaces $$[a,b],(a,b),(a,b],[a,b),(-\infty,a),(a,\infty)\ and\ \{a\}\ for\ a,b\in \mathbb{R}$$ with its induced topology is polish space.

Polish spaces:A topological space is said to be Polish space if it is separable and completely metrizable.

I know $$\mathbb{Q}\cap [a,b] \subseteq [a,b]$$ and is dense in $$[a,b]$$ and countable so $$[a,b]$$ is separable and the topology induced by $$\mathbb{R}$$ is same the topology induced by the euclidean metric on $$[a,b]$$.So it is metrizable.And since closed subset of complete metric space is complete implies $$[a,b]$$ is also complete.So,our $$[a,b]$$ with euclidean metric is separable and completely metrizable hence polish space.

But I am having doubt proving for other spaces: 1)$$(a,b)$$.This space may be metrizable,separable.But complete?? Suppose Take example of $$(0,1)$$.I have a sequence $$\{\frac{1}{n}\}\subseteq (0,1)$$This is Cauchy and converges to zero.But since the $$0\notin (0,1)$$ the sequence is not convergent in $$(0,1)$$.So doesn't it imply $$(0,1)$$ is not complete metric space and hence $$(a,b)$$.So how $$(a,b)$$ is polish space??

• See, the definition says "metrizable"? So if you can come up with a different metric (inducing the same topology) which is complete, that will do. Commented Mar 14, 2021 at 14:07

$$(a,b)$$ can be given another metric that is complete and induces the same topology.
One such metric can come from a homeomorphism $$f: (a,b) \to \Bbb R$$, where we introduce the new metric $$d(a,a')= |f(a) - f(a')|$$ and makes $$f$$ an isometry from $$((a,b), d)$$ to $$(\Bbb R, d_e)$$.
In general, $$X$$ is Polish iff it can be embedded as a $$G_\delta$$ (countable intresection of open sets) of a complete separable metric space. So open sets of $$\Bbb R$$ and the irrationals $$\Bbb P$$ are also examples of such spaces.