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I solved some short questions and I want to know if they are correct:

  1. Every subset of a topological space is either open or closed.

Answer: Wrong. Let $X\subseteq X$ is a subspace and $X$ is open and closed in $X$.

  1. An open set is not closed.

Answer: Wrong. $\emptyset$ is closed and open.

  1. Compact subsets of a Hausdorff space are closed.

Answer: True. $A\subset X$ compact, $X$ Hausdorff $\implies$ $A$ is closed in $X$

  1. If a subset of a topological space has an cover of open sets, then it's compact.

Answer: Wrong. $A\subset X$ is compact iff every cover of open sets has a finite subcover.

  1. Surfaces with the same Euler-characteristics are homeomorph.

Answer: Wrong. It is true that if $F_1\simeq F_2\implies\chi(F_1)=\chi(F_2)$ but not $\chi(F_1)=\chi(F_2)\implies F_1\simeq F_2 $.

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    All your answers are fine. – 2017-02-07
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    Your answer to (1) is questionable (the statement is indeed wrong, but not for the reason you cite). I think that "or" doesn't mean "exclusively one or the other but not both". Since a set that is both open and closed is of course open, $X$ is either open or closed. I believe they want you to demonstrate if there could be a space with a set that is neither open nor closed. – 2017-02-07
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    While your actual answers are all correct, what you've written after them for (3) through (5) are not quite formal justifications for the answers. For (3) it would mean a proof; for (4), it would be an example of a noncompact subset with a cover of open sets. Of course, if it was not your intention to give full justifications, then this is no problem. – 2017-02-07
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    Thank you everybody for the input. – 2017-02-07
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    Every subset of $X$ has a cover by open sets, the most simple of which is just $\{X\}$ itself. Then 4 would make compactness a pretty void notion. – 2017-02-08

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For the first question, I think you should say that a set such as $(0,1]\subset \mathbb{R}$ is neither open nor closed. That is, the "or" in the question should be interpreted as being the inclusive "or." Also, can you come up with counterexamples for numbers $4$,$5$?

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    Answer 4: the set $\mathbb{N}$ can be covered with $\bigcup\limits_{i=0}^\infty [i-\epsilon,i+\epsilon]$ but this has no finite subcover. Answer 5: Torus and Möbius strip have same Euler-characteristics but are not homeomorphic. – 2017-02-07
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    For 5, does your definition of surface allow for surfaces with boundary? Some authors make a distinction, in which case you should find a different counterexample with 2 surfaces with empty boundary. – 2017-02-07