# Thread: help me in Prove

1. ## help me in Prove

Prove

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

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

2. For part two, define a sequence of sets in the following way: $\displaystyle A_0=X, A_1=f(A_0), A_2=f(A_1), A_3=f(A_2),\ldots$. Since $\displaystyle f:X\to X$ is continuous and takes a compact space to a Hausdorff space, the closed map lemma applies; we can say that each $\displaystyle A_i$ is a closed set. Let $\displaystyle A=\bigcap A_i$. Note that $\displaystyle A$ is closed and $\displaystyle f(A)=A$. To complete the proof, it necessary to show that $\displaystyle A$ 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) $\displaystyle X$ is a regular space, $\displaystyle A$ compact set and $\displaystyle B$ closed set,
$\displaystyle A \bigcap B = {\O}$
Then there is open sets U and V exist, such that
$\displaystyle A\subseteq U, \ B \subseteq V,\ A \bigcap B = {\O}$
Oh come on!

Hint:

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

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

Hint:

Spoiler:

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