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Compact sets

Siron

Active member
Jan 28, 2012
150
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!
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
We know that the intersection of closed sets are closed and the intersection of bounded sets are bounded. Therefore, the intersection of closed and bounded sets are closed and bounded, i.e. compact.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
We know that the intersection of closed sets are closed and the intersection of bounded sets are bounded. Therefore, the intersection of closed and bounded sets are closed and bounded, i.e. compact.
what does "bounded" mean in the absence of a metric?

(will post more later)
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
what does "bounded" mean in the absence of a metric?

That is a very good point.
But it is true that the intersection of two closed sets is closed.
It is a theorem: In a compact Hausdorff space a subset is compact if and only it is closed.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
i had some trouble coming up with a counter-example. but here it is:

let $X = [-1,1] \times \{0,1\}$, with the topology:

$\mathcal{U} = \{U \times V: U \in \mathcal{S}, V \in \mathcal{T}\}$

where $\mathcal{S}$ is the standard (relative) topology on [-1,1] and $\mathcal{T} = \{ \emptyset, \{0,1\}\}$ (the indiscrete topology on {0,1}).

let $A = [-1,1] \times \{0\}$, $B = ([-1,0) \times \{0\}) \cup ([0,1] \times \{1\})$

it should be clear that $X$ is not Hausdorff, for there are no disjoint neighborhoods of $(0,0)$ and $(0,1)$.

it is clear that $A$ is compact, but it is not so clear that $B$ is compact.

note, however that any open cover $\mathcal{O}$ of $B$ must also cover:

$B' = ([-1,0) \times \{0,1\}) \cup ([0,1] \times\{0,1\}) = X$ and $X$ is certainly compact (since $[-1,1]$ is compact in $\mathcal{S}$ and $\{0,1\}$ is finite).

now $A \cap B = [-1,0) \times \{0\}$ and if $\mathcal{O}'$ is the open cover:

$\{U_n = [-1,\frac{-1}{n}) \times \{0,1\}\}$

we have no finite sub-cover, so $A \cap B$ is not compact.

(will write more later).
 

Ubistvo

New member
Dec 19, 2012
10
Here's why $\mathbb Q$ is not locally compact:

Suppose $\mathbb Q$ is locally compact, then there exists $K\subset \mathbb Q$ compact which contains a neighborhood, say $(a,b)\cap\mathbb Q$ of $0.$ Let $x$ be a irrational element of $(a,b),$ so there exists a Cauchy sequence $x_i$ that converges to $x$ such that for each $x_i\in\mathbb Q\cap (a,b),$ so the only limit point of the sequences $x_i$ is $x,$ but $x\not\in\mathbb Q,$ so $\{x_i\}$ has no limit points in $\mathbb Q$ and then not in $K,$ and that means that $K$ can't be compact, contradiction.