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Hi, can someone help me with the following questions.

1. Let $X$ be a Hausdorff space and suppose $A, B \subseteq X$ are compact subsets. Then $A \cap B$ is compact. What if $X$ is not Hausdorff?

In Hausdorff spaces every compact subset is closed so $A$ and $B$ are closed, thus I wanted to prove that $A\cap B$ is closed and then $A \cap B$ would be compact (because $X$ is compact). However $\mbox{cl}(A) \cap \ \mbox{cl}(B) \ = \mbox{cl}(A \cap B)$ is not always true so I thought maybe in Hausdorff spaces it is, but I can't prove it.

2. Let $(X,\mathcal{T})$ and $(Y,\mathcal{S})$ be topological spaces and suppose is continu. Let $f: X \to Y$ be continu.

Prove:

2.1 Suppose $f$ is a closed surjection, suppose $\forall y \in Y: f^{-1}(y)$ is compact and suppose $X$ is Hausdorff. Prove $Y$ is Hausdorff.

Attempt:

Let $X$ be Hausdorff and suppose $f^{-1}(y)$ is compact. We can find disjoint neighbourhouds, say $W_y$ for $f^{-1}(y)$ and $V_{y}$ for $x$ where $x \notin f^{-1}(y)$, i.e $W_{y} \cap V_{y}= \emptyset$. The problem is that $ \emptyset = f(V_{y} \cap W_{y}) \subset f(V_{y}) \cap f(W_{y})$ thus I can't immediately find two disjoint neighbourhoods for $f(x)$ and $y$. How do I solve this problem?

3. Does someone has a proof for the fact that $\mathbb{Q}$ is not locally compact?

Thanks in advance!

1. Let $X$ be a Hausdorff space and suppose $A, B \subseteq X$ are compact subsets. Then $A \cap B$ is compact. What if $X$ is not Hausdorff?

In Hausdorff spaces every compact subset is closed so $A$ and $B$ are closed, thus I wanted to prove that $A\cap B$ is closed and then $A \cap B$ would be compact (because $X$ is compact). However $\mbox{cl}(A) \cap \ \mbox{cl}(B) \ = \mbox{cl}(A \cap B)$ is not always true so I thought maybe in Hausdorff spaces it is, but I can't prove it.

2. Let $(X,\mathcal{T})$ and $(Y,\mathcal{S})$ be topological spaces and suppose is continu. Let $f: X \to Y$ be continu.

Prove:

2.1 Suppose $f$ is a closed surjection, suppose $\forall y \in Y: f^{-1}(y)$ is compact and suppose $X$ is Hausdorff. Prove $Y$ is Hausdorff.

Attempt:

Let $X$ be Hausdorff and suppose $f^{-1}(y)$ is compact. We can find disjoint neighbourhouds, say $W_y$ for $f^{-1}(y)$ and $V_{y}$ for $x$ where $x \notin f^{-1}(y)$, i.e $W_{y} \cap V_{y}= \emptyset$. The problem is that $ \emptyset = f(V_{y} \cap W_{y}) \subset f(V_{y}) \cap f(W_{y})$ thus I can't immediately find two disjoint neighbourhoods for $f(x)$ and $y$. How do I solve this problem?

3. Does someone has a proof for the fact that $\mathbb{Q}$ is not locally compact?

Thanks in advance!

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