# Is completeness of a metric space is a homeomorphic property?

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.

• I would call a property “topological”, not “homeomorphic”, if it is invariant under homeomorphisms. – pyon Jul 5 '19 at 8:45
• @pyon the question from the assignment that I have mentions "homeomorphic", i.e, the question word for word. So I'm a bit lost. – Brian I. Hunds Jul 5 '19 at 8:46
• It is not al all clear what you want to know. – Paul Frost Jul 5 '19 at 8:47
• "In one of those metrics, (0,1) is complete, but in the other it is not, though the two metrics generate the same topology." This is precisely enough to prove that completeness is not invariant under homeomorphisms. In this case the identity function between your two metric spaces is a homeomorphism (precisely because the topologies are the same). So they are homeomorphic, but one is complete and the other is not. – preferred_anon Jul 5 '19 at 8:56
• It's not saved via homeomorphisms, but it is saved via uniform homeomorphisms (i. e. we demand bi-uniform continuity instead of continuity). – Jakobian Jul 5 '19 at 8:57

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