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I have 2 topology questions with respect to the standard topology:

1) Is it true or false, justify the answer: $\Bbb R \setminus \Bbb Q$ is compact.

2) Prove that $[0, 1] × [0, 1]$ and $(0, 1) × (0, 1)$ are not homeomorphic using compactness.


Here is what I have:

1) $\Bbb R\setminus \Bbb Q$ is a set of irrational numbers, which are not bounded. In $\Bbb R$, compactness means being closed and bounded. Thus, $\Bbb R\setminus \Bbb Q$ is not compact.

2) For the sake of contradiction, assume they are homeomorphic. $[0, 1] × [0, 1]$ is compact because it's closed and bounded, thus $(0, 1) × (0, 1)$ is compact and it's closed as well as bounded. However, $(0, 1) × (0, 1)$ is not closed because finite product of open sets are still open(is this correct?)

Thank you in advance for the help!

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2 Answers 2

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1) Is it true or false, justify the answer: $R $\ $Q$ is compact.

It is false, because $R$ \ $ Q$ is not bounded. Compact sets are bounded.

2) Prove that $[0, 1] × [0, 1]$ and $(0, 1) × (0, 1)$ are not homeomorphic using compactness.

$[0, 1] × [0, 1]$ is compact because it is closed and bounded.

$(0, 1) × (0, 1)$ is not compact because it is not closed.

Since compactness is a topological property, a compact set and a non-compact set are not homeomorphic.

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Rather, you can claim that $(0,1)\times(0,1)$ is not compact by considering the open covering $\{(1/n,1)\times(1/n,1): n=1,2,...\}$.

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  • $\begingroup$ Got it! Did I get 1) right? $\endgroup$
    – Liz
    Mar 7, 2018 at 2:57
  • $\begingroup$ Okay, it goes through. $\endgroup$
    – user284331
    Mar 7, 2018 at 2:58

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