What does metrizablity means Let $(X,T)$ be a topological space, and let $f$ be a homeomorphism form $(X,T)$ to any metric space. Than we say that $(X,T)$ is metrizabile. This means that X can be equipped with a metric and can be seen as a metric space. My question is does it mean that $(X,T)$ is a metric space as well, so does there exist a metric in X s.t. the topology be the same as it was $(T)$? If so, does that mean that $(X,T)$ and $(X,d)$ (where $d$ would be that metric) are identically only in topological view or they would be the same in any other aspect (i.e. does it mean that only topological properties of metric spaces inherit to $(X,T)$ or any other property is the same for both spaces)?
 A: Yes, a space is metrizable if the topology can be represented as a metric topology. However, multiple metrics may generate the same topology. That is why the space is called metrizable as long as no metric has been fixed.
A: Suppose $(X, \mathcal{T})$ is a topological space and $X$ is metrizable in your sense that there exists a metric space $(Y,d)$ and a homeomorphism $f$ between $X$ and $Y$.
This means that topologically we can identify $X$ and $Y$: any topological property that $Y$ has, $X$ has as well, and any topological property that $X$ has, $Y$ has as well. For instance, we know that all metric spaces are paracompact so $Y$ is paracompact, and this is a topological property of $Y$, so $X$ is paracompact as well. The same can be said for $X$ being perfectly normal and first countable (examples of topological properties that metric spaces have, by virtue of having a topology induced by a metric). 
In fact, we can define a metric $d_f$ on the set $X$ by $d_f(x,y) = d(f(x), f(y))$ (this is clearly a metric, using that $f$ is a bijection and $d$ a metric on $Y$) and one can check that $B_{d_f}(x, r) = f^{-1}[B_d(f(x), r)]$, so open balls in $X$ under $d_f$ are open in $\mathcal{T}$ (as $f$ is continuous and open balls are open in $Y$ by definition), and some more checking shows that $\mathcal{T}$ is indeed exactly the topology induced by $d_f$ on $X$.
On the other hand, there are properties of $Y$ that are formulated in terms of the metric $d$, like $Y$ is complete. As this property is not purely topological, we cannot say that $(X, \mathcal{T})$ is complete. But we can say that if $d$ is a complete metric, then $d_f$ is a complete metric on $X$ as well (clear from the definitions). We can say that $X$ is completely metrizable (i.e. there is a complete metric that induces the topology). But if we give $X$ another metric such that $X$ is metrizable, it need not be the case that $X$ is complete in that metric: the completeness does depend on the specific metric that is chosen, and this is by no means unique. 
