# Proper dense open subset of X

$$X$$ 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.

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?

• I think he means the difference between sets. – InsideOut May 27 at 4:48
• yes , difference between sets – Arindam basak May 27 at 4:50

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.

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

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

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