Two problems: When a countinuous bijection is a homeomorphism? Possible cardinalities of Hamel bases?

Question

Let $X$ and $Y$ be topological spaces and let $f : X\rightarrow Y$ be a continuous bijection. Under which of the following conditions will $f$ be a homeomorphism?
(a) $X$ and $Y$ are complete metric spaces.
(b) $X$ and $Y$ are Banach spaces and $f$ is linear.
(c) $X$ is a compact topological space and $Y$ is Hausdorff.
Let $V$ be a complete normed linear space and let $B$ be a basis for $V$ as a vector space. Pick out the true statements:
(a) $B$ can be a finite set.
(b) $B$ can be a countably infinite set.
(c) If $B$ is infinite, then it must be an uncountable set.

It would be helpful to know why you think this, along with whatever other work you may have done. It would also be helpful in the future to post different questions as different questions with descriptive titles. That way, people can locate similar questions more easily, instead of wading through a sea of "problem on topology" questions.
Continuous Bijection from Compact to Hausdorff is Homeomorphism at ProofWiki.
A continuous bijection from a complete metric space to a complete metric space is not necessarily a homeomorphism. There is, for instance, a continuous bijection from $\Bbb R$ to a space in the shape of figure $8$. (We may also desribe the latter as $S^1\vee S^1$.)
@poton: when you have several unrelated questions, you should post them as separate questions.

tomasz · Accepted Answer · 2012-08-09 14:49:58Z

When you have several mostly unrelated questions, don't post them as one question.

Adding up partial answers from the comments, and some of my own:

1a. is false, because it is easy to see a continuous bijection between $\mathbf R$ and a figure-eight in $\mathbf R^2$, which are not homeomorphic at all.
1b. is true, because the open mapping theorem applies in this case, so a continuous linear bijection is open and hence homeomorphic
1c. is true, and is actually a basic result in general topology which can be found in many books as well as the internet, for example here.
2a. is obviously true ($\mathbf C$ is finite dimensional and complete...)
2b. is false, because a proper subspace of any normed space is closed nowhere dense, and the span of a countable subset of a normed space is the union of the spans of its finite subsets, so a countable union of closed nowhere dense sets, and hence not the entire space by Baire category theorem.
2c. is true because 2b. is false.

Byron Schmuland · Answer 2 · 2012-08-13 03:05:34Z

a) thanx to thomasz, for figure 8 counter example. so False.

b) true, by open mapping theorem. surjective maps are open.

c) continuous bijective map from compact space to Hausdorff space is homeomorphism, to prove this enough to show that $f$ is closed map, take a closed set $A$, then it is compact also (why?), consider its image $f(A)$ it is compact (why?), and compact subset of a Hausdorff space is closed (need proof?), so $f$ is closed map, so $f$ is homeomorphism.
(a) true, as you can take a finite dimensional space

(b) is false by Baire Category theorem.

(c) is true by Baire Category theorem.

$f(x)=x^3$ is a homeomorphism from $\mathbb R$ to $\mathbb R$ since $g(x)=\sqrt[3]{x}$ is continuous.
oops sorry, I thought Diffeomorphism :P, ok I am thinking and changing my answer.
For your 1. a), you can just take the map Dejan alluded to in the above comments.

2 Answers