Let X be the cartesian product of $\displaystyle \mathbb{R}$ 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?

Printable View

- Oct 26th 2009, 08:57 PMdori1123intersection of compact sets
Let X be the cartesian product of $\displaystyle \mathbb{R}$ 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? - Oct 27th 2009, 04:56 AMHallsofIvy
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.

- Oct 27th 2009, 05:13 AMtonio
- Oct 27th 2009, 10:43 AMEnrique2
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 $\displaystyle \{0\}$ or $\displaystyle \{1\}$ are open nor closed in $\displaystyle \{0,1\}$ endowed with the trivial topology.

I guess these two subsets

$\displaystyle K_1:=[-1,1]\times \{1\}$

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

are compact, and the intersection

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

I suggest proving the compactness of these subsets - Oct 27th 2009, 04:03 PMJose27
- Oct 27th 2009, 04:25 PMEnrique2
I believe both are compacts. $\displaystyle K_1$ is homeomorphic to $\displaystyle [-1,1]$. Any sequence or net $\displaystyle (x_i,y_i)_i$ in this space is convergent if and only if $\displaystyle (x_i)_i$ is. Notice that $\displaystyle (y_i)_i$ is always convergent to both 0 and 1. Hence, roughly speaking,

$\displaystyle K\times $"anything" is compact iff $\displaystyle K$ is compact in $\displaystyle \mathbb{R}$. The argument of an open recovering

$\displaystyle V_i\times W_i$ also work because $\displaystyle W_i$ is always the whole space. I think $\displaystyle K\subset X$ is compact if and only if the projection $\displaystyle p_1(K)$ is compact in $\displaystyle \mathbb{R}$.

Anyway I prefer people to check this, non Hausdorff spaces are not familiar to me, but I like the problem. - Oct 27th 2009, 04:40 PMaliceinwonderland
I think this is a nice example. To see $\displaystyle K_1$ is compact and $\displaystyle K_1\cap K_2$ is not compact are rather straightforward.

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