1. help me in Prove

Prove

1) $X$ is a regular space, $A$ compact set and $B$ closed set,
$A \bigcap B = {\O}$
Then there is open sets U and V exist, such that
$A\subseteq U, \ B \subseteq V,\ A \bigcap B = {\O}$

2) $X$ is compact and hausdorff space and $f : X \rightarrow X$ is continuous function
prove there is a closed set non empty such that $f(A)= A$

2. For part two, define a sequence of sets in the following way: $A_0=X, A_1=f(A_0), A_2=f(A_1), A_3=f(A_2),\ldots$. Since $f:X\to X$ is continuous and takes a compact space to a Hausdorff space, the closed map lemma applies; we can say that each $A_i$ is a closed set. Let $A=\bigcap A_i$. Note that $A$ is closed and $f(A)=A$. To complete the proof, it necessary to show that $A$ is nonempty. I can't see how to do that at the moment, so hopefully somebody else can fill in the gap.

I hope that this gives you some ideas, at any rate.

3. Originally Posted by nice rose
Prove

1) $X$ is a regular space, $A$ compact set and $B$ closed set,
$A \bigcap B = {\O}$
Then there is open sets U and V exist, such that
$A\subseteq U, \ B \subseteq V,\ A \bigcap B = {\O}$
Oh come on!

Hint:

Spoiler:
For each $a\in A$ there are disjoint neighborhood $U_a,V_a$ with $a\in U_a$ and $B\subseteq V_a$. Cover $A$ with the set of all the $U_a$'s and procure a finite subcover. What then?

2) $X$ is compact and hausdorff space and $f : X \rightarrow X$ is continuous function
prove there is a closed set non empty such that $f(A)= A$
Let's see some work

Hint:

Spoiler:

What happens if $K=\left\{x\in X:f(x)\ne x\right\}=X$?