Topology: Example of a compact set but its closure not compact Can anyone gives me an example of a compact subset such that its closure is not compact please? Thank you.
 A: The other answers are fantastic. I present mine since I find it to be a useful "test case" to keep in mind in other situations, especially for statements about limits.
Consider $\mathbb{R}$ with the following topology: $U\in\tau$ if and only if $$U = (a,\infty),\hspace{2ex}a\in[-\infty,\infty],$$
where $(\infty,\infty)$ is understood to be the empty set and $(-\infty,\infty) = \mathbb{R}$.
Any singleton $\{x\}\subset \mathbb{R}$ is compact: given an open cover $\{U_\alpha\}_{\alpha\in A}$, take any one element of the cover; by definition it contains $\{x\}$, so there is a finite subcover consisting of just one set.
The closure of $\{x\}$ is $(-\infty, x]$. This set is not compact: the open cover consisting of all sets of the form $(a,\infty)$ for $a>-\infty$ has no finite subcover.
A: Write $\tau$ for the standard topology on $\Bbb R$. Consider now $\tau_0=\{U\cup\{0\}\mid U\in\tau\}\cup\{\varnothing\}$.
It's not hard to verify that $\tau_0$ is a topology on $\Bbb R$. It's also not hard to see that $\overline{\{0\}}=\Bbb R$. However one can easily engineer an open cover without a finite subcover.
A: I think the simplest example would probably be the "particular point topology" on an infinite set. 

Let $X = \{p\} \cup \mathbb{N}$, topologized so that the nonempty open sets are the sets containing $p$.

The singleton $\{p\}$ is compact (finite sets are always are). However, $\overline{ \{p\}} =X$, and $X$ is not compact because the open cover $X = \bigcup_{n\in \mathbb{N} } \{p , n\}$ has no finite subcover. 
A: Here is an example (using the formalism of affine schemes) illustrating Elden's excellent idea.  
Let $k$ be a field and $\mathbb A^\infty_k$ be the affine scheme associated to the polynomial ring over $k$ in infinitely many variables: $$\mathbb A^\infty_k=\operatorname {Spec} (k[T_n|n\in \mathbb N])$$  
For our  example we will consider the open subspace  $$X=\mathbb A^\infty_k\setminus \{\mathfrak m\}$$ where $\mathfrak m$ is the maximal ideal $\mathfrak m=(T_0 ,...,T_n,..)\subset k[T_n|n\in \mathbb N]$
That subspace $X$  has the  point $\eta=(0)$ corresponding to the zero ideal as its generic point, meaning that  the singleton set $\{\eta\}$ is  dense in $X$ : $\overline{ \{\eta\}}=X$ .
Now  $\{\eta\}$ is certainly compact but its closure $X$ is not quasi-compact:
Indeed, $X$ is covered by the family of principal open subsets $(D(T_n))_{n\in \mathbb N}$ but a finite union $\bigcup^N_{n=0} D(T_n)$ can never  cover $X$ because for the prime ideal $(T_0,...,T_N)\in X$ we have  $(T_0,...,T_N)\notin \bigcup^N_{n=0} D(T_n)$.
A: There are no Hausdorff examples; T$_0$ examples are trivial; the following T$_1$ example is only slightly less trivial.
Let $A,B$ be disjoint infinite sets, and let $X=A\cup B$ with the topology $$\tau=\{\emptyset\}\cup\{U\subseteq X:A\setminus U\text{ is finite}\}.$$ Then $X$ is a T$_1$-space, $A$ is a compact subspace of $X$, and $\overline A=X$ is not compact.
A: Consider $\Bbb R$ with topology generated by basis $B=\{(-\infty,a):a \in\Bbb R\}$ then the set $K=\{1/n : n \in\Bbb N\}$ is compact in this topology but its closure $[0, \infty)$ is not compact. Is this answer correct?
