Compactness of Topological Spaces The only one that I have been able to get at is that (c) is not compact since it is not closed/bounded by Heine-Borel Theorem. Any thoughts on how to approach the others? I understand that, in order to prove a space is not compact, I must provide a counterexample subcover (for some open cover of the space) that is not finite, correct?
Decide which of the following spaces are compact.
a.
The plane $R^{2}$ with the topology which has for a subbasis $S=\{U \:\vert \: U=R^{2}-L, \: where \: L \: is \: any \: straight \: line\}$.
b.
The plane $R^{2}$ where an open set is any set of the form $R^{2}-C$, where C contains at most countably many points of $R^{2}$, and $\emptyset$ is open.
Solution:
The plane $R^{2}$ where an open set is any set of the form $R^{2}-C$, where C contains at most countably many points of $R^{2}$, and $\emptyset$ is open.
Define the space to be T. T is not compact since it is an infinite subset of $R$, i.e. it contains a countable set of distinct points $a_{n}$, $n \in N$, since T has infinitely many elements. 
$U_{n} := R \setminus \: \{a_{k} \: | \: k \geq n\}$. Note that $R \setminus U_{n} = \{a_{k} \: | \: k \geq n\}$ is countable, so $U_{n}$ is open.
$\cup_{n \geq 1} U_{n}=R \rightarrow$ this collection is an open cover of T. Let the T be compact s.t. $T \subseteq U_{1} \cup ... \cup U_{N}$ for some natural N. However, $U_{1} \cup ... \cup U_{N}=R \setminus \: \{a_{k} \: | \: k \geq N\}=U_{N}$, so $a_{N} \in T \subseteq U_{1} \cup ... \cup U_{N}$ and $a_{N} \notin U_{1} \cup ... \cup U_{N}$. This is a contradiction, so the space T is not compact.
c.
The subspace of rational numbers in the usual space of real numbers.
d.
The space in which X is the set of all functions from the set R of real numbers into R, making no assumption about the continuity of these functions. To define the topology on X, specify an open neighborhood system. Suppose f is any element of X. Let F be any finite subset of R and p be any positive real number. Define $U(f, F, p)=\{g \in X \: \vert \: \vert \: g(x)-f(x) \: \vert \: < p \: \forall \: x \in F\}$. Let $N_{f}$ be the set of all $U(f, F, p)$ for all finite subsets F of R and all positive numbers p. Note that for a given f, $U(f, F, p)$ depends on both F and p; hence $U(f, F, p)$ is not a p-neighborhood in the metric sense.
 A: This is a partial answer.
Example (b) is not compact. Pick any countably infinite subset $C$ of $\mathbb{R}^2$ and label the elements as $\{c_1, c_2, ...\}$. Then consider the sets $C_i=\{c_{i+1}, c_{i+2}, ...\}$ and let $C_0=C$. The collection $\{\mathbb{R}^2-C_i\}$ is a cover of $\mathbb{R}^2$ which clearly has no finite subcover, as any finite subcover necessarily excludes countably many points.
A: Example (a) is compact. By Alexander theroem, it is enough to consider covers by open sets from the subbasis. If we take one of the sets, we have covered everything but one line. If there is a complement of a parallel line in the cover then we are done, if not then there is a complement of a crossing line and we have covered everything but one point by two sets. We just take another set covering the point. So actually any cover by subbasic subssets has a subcover consisting of three sets.
As for (d), correct me if I'm wrong, but isn't this just product topology on $\mathbb{R}^\mathbb{R}$? In this case it is not compact, because $\mathbb{R}$ is its continuous image which is not compact.
