# What about the compactness and connectedness of $[0,1]$ in this topology?

If the set $\mathbf{R}$ of all real numbers has the topology consisting of all sets $A$ such that $\mathbf{R} \setminus A$ is either countable or all of $\mathbf{R}$. What can we say about compactness and connectedness of the closed interval $[0,1]$ in this topology?

Hint: (1) For closed, you have to answer: Is $\mathbb R\setminus[0,1]$ countable or the whole of $\mathbb R$?

(2) For compact: Let $A = \{a_n \mid n\in \mathbb N\} \subseteq [0,1]$ countable infinite, let $O_n = \mathbb R \setminus \{a_k \mid k \ge n\}$. What can you say about the sets $O_n$?

For compactness, pick some countable subset of $[0,1]$ and call it $P$. Enumerate the elements in $P$ so that $P=\{p_0,p_1,p_2,\ldots\}$. Now let $P_s=P\cup\{p_0,\ldots p_{s-1}\}=\{p_s,p_{s+1},\ldots\}$. We now take our open cover $\mathcal{U}$ to be the collection $\mathcal{U}=\{P_s\mid s\in\mathbb{N}\}$. Clearly $\mathcal{U}$ is a cover of $[0,1]$ as every point in $[0,1]$ is contained in some $P_s$. However, it is also clear that no finite subset of $\mathcal{U}$ can be an open cover of $[0,1]$ because any finite union of elements in $\mathcal{U}$ will still be missing an infinite number of elements from $[0,1]$. Hence $[0,1]$ is not compact in this topology.

For connectedness, note that $[0,1]\setminus P$ and $[0,1]\setminus P'$ for some countable $P$ and $P'$ must have non-empty intersection because $[0,1]\setminus (P\cup P')$ is an uncountable set less a countable set and so is still uncountable. As every open set in the topology is of the form $\mathbb{R}\setminus P$, there is no way to partition $[0,1]$ in to two disjoint open subsets by the above arguement. Hence $[0,1]$ is connected.

$X\subseteq\mathbb{R}$ be finite then it is compact in this topology (why?)

$Y\subseteq\mathbb{R}$ be infinite then it is not compact as it contains a countable set of distinct points $x_n$ define $U_n=\mathbb{R}\setminus \{x_k:k\ge n\}$, Clearly $U_n$'s are open according to your topology and forms an open cover of $\mathbb{R}$ so it is a cover for $[0,1]$

• Finite subsets are compact in any topological space. Sep 20 '13 at 11:23