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$X$ be a topological space and $U$ be a proper dense open subset of $X$. Then pick the correct statement from the following:

  1. If $X$ is connected then $U$ is connected.
  2. If $X$ is compact then $U$ is compact.
  3. If $X\setminus U$ is compact then $X$ is compact.
  4. If $X$ is compact then $X\setminus U$ is compact.

Answer:

My approach is, while $U$ is open and dense subset therefore if $X$ is compact and $U$ is open then $X\setminus U$ is closed subset of X and also it is non-empty, hence $X\setminus U$ is compact.

So option 4 is correct.But I don't know about other options. What to do?

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  • $\begingroup$ I think he means the difference between sets. $\endgroup$ – InsideOut May 27 at 4:48
  • $\begingroup$ yes , difference between sets $\endgroup$ – Arindam basak May 27 at 4:50
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  1. $X=[-1,1]$ (usual topology), is connected and its subspace $U=\mathbb{R}\setminus \{0\}$ is open and dense and disconnected. False.

  2. As 1., note that a compact subset of $[-1,1]$ is closed and $U$ is not.

  3. As 1. $X\setminus U=\{0\}$ is compact.

  4. True, as $X\setminus U$ is closed in $X$ ($U$ is open) and so compact too when $X$ is.

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1) can't be, because if we take $X = \mathbb{R}$, $U= \mathbb{R} - \{0\}$

Then $X$ is a conected set, $U$ a open proper dense subset, but $U$ is not conected

2) can't be, because if we take $X=[0,1]$ with relative topology of $\mathbb{R}$ and $U = (0,1)$

Then $X$ is a compact set, $U$ a subset open dense of $X$ but $U$ is not a compact set.

3) can't be, because if we take $X=\mathbb{R}$ and $U = \mathbb{R} - \{0\}$

Then $X-U = \{0\}$ is a compact set, but $X$ is not compact

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1,2. $X=[-1,1]$, $U= X \setminus \{0\}$.

3.$X=(-1,1)$, $U= X \setminus \{0\}$.

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