Is completeness of a metric space is a homeomorphic property?

Question

A property of a topological space is called a homeomorphic property if it remains invariant under homeomorphisms

Is boundedness of a subspace of a metric space is a homeomorphic porperty?

For the first question, I can prove so far as to show that completeness is not a Topological property as if you consider (0,1), then one metric is the ordinary Euclidean metric on (0,1); a second metric is what you get when you pull the Euclidean metric on R back to (0,1) via a homeomorphism.

In one of those metrics, (0,1) is complete, but in the other it is not, though the two metrics generate the same topology. But how do I show go about showing that comepleteness remains invariant under homeomorphisms (or not)?

For the second question I am not sure where to begin as total boundedness is a not a topological property but I cannot say about boundedness in general. And it talks of metric space in the question, so this line of reasoning won't work at all I guess.

Answer 1

Since $(0,1)$ is bounded and homeomorphic to $\mathbb R$, which is unbounded, the answer is negative.

Answer 2

A topological space is a pair $(X,T)$ where $T$ is a topology on the set $X$. A metric space is a triplet $(X,T,d)$ where $d$ is a metric on the set $X,$ and $T$ is the topology on $X$ generated by the base of all open $d$-balls. A metrizable space is a topological space $(X,T)$ such that $(X,T,d)$ is a metric space for some metric $d$ on $X.$ A completely metrizable space is a topological space $(X,T)$ such that $(X,T,d)$ is a metric space for some complete metric $d$ on $X$.

It is very common, and generally acceptable, to speak of any of these as "the space $X$". But this should not be done if you are talking about two topologies or two metrics on $X$.

A property $p(A)$ that a space $A$ may, or may not have, is called a topological property iff $p(A)\iff p(B)$ whenever $A,B$ are homeomorphic. Examples: 1. Compact 2.Connected 3. Path-connected. 4. Metrizable. 5.Completely metrizable. These are topological properties.

Being bounded is NOT a topological property. If $d$ is a metric on $X$ then the metric $e(x,y)=\min (1,d(x,y)\,)$ on $X$ generates the same topology on $X$ that $d$ does. And $e$ is a bounded metric but $d$ might not be.