Counterexample to the intersection of compact sets being compact.

Question

I want to give a counterexample to the intersection of compact sets in a general topological space being compact. I am aware of this counterexample, but I was hoping to obtain one of the following sort:

Two results:

$1)$ A net in a topological space $(X,\mathcal{T})$ has a unique limit $\iff$ $X$ is Hausdorff.

$2)$ In any topological space $(X,\mathcal{T})$, if $x_n\rightarrow x$ then the set $\{x_n:n\in\mathbb{N}\}\cup\{x\}$ is compact.

We are assuming throughout that $X$ is uncountable. I was hoping to find some suitable topology which is not Hausdorff, then we know that there should exist a sequence which converges to two distinct points in the sense of net convergence say $x$ and $y$. (1)

We simply take the intersection $\left(\{x_n:n\in\mathbb{N}\}\cup\{x\}\right)\bigcap\left(\{x_n:n\in\mathbb{N}\}\cup\{y\}\right)=\{x_n:n\in\mathbb{N}\}$, now we only need to find an open cover which has no finite subcover. It would be enough to show that we have an open cover where all of the elements of $\{x_n:n\in\mathbb{N}\}$ are contained in disjoint open sets.

My initial hunches of suitable topologies were possibly the co-countable or discrete topologies.

Unfortunately, it's clear that the discrete topology will always be Hausdorff, so the example won't get off the ground.

It might be possible to construct open sets $U_i$ in the co-countable topology that only include $x_i$ and exclude all other entries of the sequence. Unfortunately, it turns out all convergent sequences in the co-countable topology are eventually constant. So this example won't get off the ground either.

Any feedback would be welcome.

(1): Do we indeed always know that in a non-Hausdorff space there exists a sequence that can converge to two separate limit points? Perhaps I have answered myself in this respect, I do not believe that the co-countable topology is Hausdorff.

So perhaps a specific example is called for!

Thanks.

PS. It seems like the topology I am hoping for needs to simultaneously act like a Hausdorff space, in the sense that we need a sufficient level of separation of the points in the space, while we explicitly state that it may not be one.

Answer 1

There is not always a sequence converging to more than one point.

Let $Y=\omega_1$, the set of countable ordinals, let $p$ and $q$ be distinct points not in $Y$, and let $X=Y\cup\{p,q\}$. $Y$ is an open subspace of $X$ with its usual order topology. Basic open nbhds of $p$ are the sets of the form $\{p\}\cup\{\xi\in Y:\xi>\eta\}$ for some $\eta\in Y$, and basic open nbhds of $q$ are defined similarly, with $q$ replacing $p$ in the definition. The sets $Y\cup\{p\}$ and $Y\cup\{q\}$ are compact Hausdorff spaces; they are in fact homeomorphic to the one-point compactification of$Y$. Their intersection, however, is $Y$, which is not compact. Finally every sequence in $Y$ is bounded in $Y$, so neither $p$ nor $q$ is the limit of a non-trivial sequence in $X$. (In fact $Y$ is an example of a sequentially compact Hausdorff space that is not compact.)

Answer 2

Let $X=(\mathbb Q \times \{0\})\; \cup \;((\mathbb R$ \ $\mathbb Q) \times \{1,2\})).$ Let $p:X\to \mathbb R$ be the projection of $X$ to its first co-ordinate. Let $T$ be the standard topology on $\mathbb R.$ Let $$B=\{(p^{-1}t) \backslash A: t\in T \land (A \text { is finite) }\}.$$ Let $X$ have the topology generated by the base $B.$ Then $X$ is a $T_1$ space (as $X$ \ $\{p\}\in B$ for $p\in X$) but not a $T_2$ space.

For $i\in \{1,2\}$ the subspace $Y_i=(\mathbb Q \times \{0\})\cup ((\mathbb R$ \ $\mathbb Q) \times \{i\}))$ is homeomorphic to $\mathbb R.$ (The projection of $Y_i$ to its first co-ordinate is a homeomorphism of $Y_i$ to $\mathbb R$).

So $Y_1\cap p^{-1}[0,1]$ and $Y_2\cap p^{-1}[0,1]$ are compact, but their intersection $([0,1]\cap \mathbb Q)\times \{0\}$ is homeomorphic to $[0,1]\cap \mathbb Q,$ which is not compact.

Let $x\in \mathbb R$ \ $\mathbb Q$ and let $(x_n)_{n\in \mathbb N}$ be a sequence in $\mathbb R$ converging to $x.$ For $n\in \mathbb N$ let $y_n\in p^{-1}\{x_n\}.$ For any $m\in \mathbb N,$ the closure of $S_m=\{y_n:n>m\}$ in the space $X$ is $S_m\cup \{(x,1),(x,2)\}.$

Remark. I thought of this space (although surely others have) for an example of a $T_1$ space with 2 completely metrizable subspaces ($Y_1$ and $Y_2$) whose intersection ($\mathbb Q \times \{0\}$) is not completely metizable. If $S$ is Hausdorff and $F$ is a countable family of completely metrizable subspaces of $S$ then $\cap F$ is completely metrizable.

Answer 3

Let set S is R + a + b (a, b - points),

Open sets is open sets in R and R+a and R+b and R + a+ b

R+a and R+b are compact sets, but it's intersection = R, in not the compact set