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.
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.