3
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $O_n$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

$X\subseteq\mathbb{R}$ 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 $[0,1]$

$\endgroup$
1
  • 1
    $\begingroup$ Finite subsets are compact in any topological space. $\endgroup$
    – Dan Rust
    Sep 20 '13 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.