intersection of compact sets

**dori1123** · October 26th 2009, 08:57 PM

Let X be the cartesian product of $\mathbb{R}$ with the usual topology and a two point set with the indiscrete topology. Find two compact subspaces of X such that their intersection is not compact.

Can I get some help please?

**HallsofIvy** · October 27th 2009, 04:56 AM

That's not possible. A compact set is closed in any topology. The intersection of two closed sets is closed in any topology. A closed subset of a compact set is compact in any topology. Therefore, the intersection of two compact sets is compact is always compact no matter what topology you have.

**tonio** · October 27th 2009, 05:13 AM

Originally Posted by HallsofIvy

That's not possible. A compact set is closed in any topology.

Only if the topology is Hausdorff, otherwise it may not be true. In remark 1.3. in A Note on Topological Properties of Non-Hausdorff Manifolds there's a nice example of a compact non-closed set. It is rather bewildering since the set is the well-known closed interval [0,1], and even more anti-intuitive: this set is Hausdorff (but the whole space, of course, isn't), yet it is a cute exercise to show that under the topology defined there this interval indeed isn't closed.

Tonio

The intersection of two closed sets is closed in any topology. A closed subset of a compact set is compact in any topology. Therefore, the intersection of two compact sets is compact is always compact no matter what topology you have.

.

**Enrique2** · October 27th 2009, 10:43 AM

Originally Posted by tonio

.

By the way, all the rest of the argument the Hallsofivy is true. Closed in compact space (i.e closed intersected compact) is always compact, in every topological space, as it is discussed in other threads. Hence this exercise needs to show that there is a non closed compact in this space. Further, they are needed two of these compacts for intersecting them, because closed and compact would give compact . I suppose the non discrete topology in {0,1} is that whose unique open sets are all the space and the empty set. Thus we have to take benefit from the fact that neither $\{0\}$ or $\{1\}$ are open nor closed in $\{0,1\}$ endowed with the trivial topology.

I guess these two subsets

$K_1:=[-1,1]\times \{1\}$
$K_2:=([-1,0] \times \{0\})\cup ((0,1]\times \{1\})$

are compact, and the intersection

$K_1\cap K_2 =(0,1]\times \{1\}$ clearly is not (is isomorphic to (0,1]).
I suggest proving the compactness of these subsets

**Jose27** · October 27th 2009, 04:03 PM

Originally Posted by Enrique2

By the way, all the rest of the argument the Hallsofivy is true. Closed in compact space (i.e closed intersected compact) is always compact, in every topological space, as it is discussed in other threads. Hence this exercise needs to show that there is a non closed compact in this space. Further, they are needed two of these compacts for intersecting them, because closed and compact would give compact . I suppose the non discrete topology in {0,1} is that whose unique open sets are all the space and the empty set. Thus we have to take benefit from the fact that neither $\{0\}$ or $\{1\}$ are open nor closed in $\{0,1\}$ endowed with the trivial topology.

I guess these two subsets

$K_1:=[-1,1]\times \{1\}$
$K_2:=([-1,0] \times \{0\})\cup ((0,1]\times \{1\})$

are compact, and the intersection

$K_1\cap K_2 =(0,1]\times \{1\}$ clearly is not (is isomorphic to (0,1]).
I suggest proving the compactness of these subsets

Is $K_1$ compact? I believe that compact subspaces in this space are exactly those of the form $K \times \{ \pm \infty \}$ where $K$ is compact in $\mathbb{R}$

**Enrique2** · October 27th 2009, 04:25 PM

Originally Posted by Jose27

Is $K_1$ compact? I believe that compact subspaces in this space are exactly those of the form $K \times \{ \pm \infty \}$ where $K$ is compact in $\mathbb{R}$

I believe both are compacts. $K_1$ is homeomorphic to $[-1,1]$ . Any sequence or net $(x_i,y_i)_i$ in this space is convergent if and only if $(x_i)_i$ is. Notice that $(y_i)_i$ is always convergent to both 0 and 1. Hence, roughly speaking,
$K\times$ "anything" is compact iff $K$ is compact in $\mathbb{R}$ . The argument of an open recovering
$V_i\times W_i$ also work because $W_i$ is always the whole space. I think $K\subset X$ is compact if and only if the projection $p_1(K)$ is compact in $\mathbb{R}$ .

Anyway I prefer people to check this, non Hausdorff spaces are not familiar to me, but I like the problem.

**aliceinwonderland** · October 27th 2009, 04:40 PM

Originally Posted by Enrique2

$K_1:=[-1,1]\times \{1\}$
$K_2:=([-1,0] \times \{0\})\cup ((0,1]\times \{1\})$

are compact, and the intersection

$K_1\cap K_2 =(0,1]\times \{1\}$ clearly is not (is isomorphic to (0,1]).

I think this is a nice example. To see $K_1$ is compact and $K_1\cap K_2$ is not compact are rather straightforward.

To check if $K_2$ is compact as Enrique2 already mentioned, any open cover for $K_2$ must cover $K_2':=([-1,0] \times \{0,1\})\cup ((0,1]\times \{0, 1\}) = [-1,1] \times \{0,1\}$ . Now we see that $K_2$ is compact.

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