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.
closed as off-topic by user147263, Tunk-Fey, Hakim, Davide Giraudo, user642796 Sep 8 '14 at 11:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Tunk-Fey, Hakim, Davide Giraudo, user642796
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.
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.