Connected and Compact Sets

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 2 November 2009

Connected Sets

A topological space is connected if $X$ satisfies the following condition. There exists open sets $A$ and $B$ of $X$ such that

$A \neq \emptyset$ and $B \neq \emptyset$ , and
$A \cup B = X$ and $A \cap B = \emptyset$ .

Let $X$ be a topological space. A subspace of $X$ is a subset $E$ of $X$ with the topology given by defining $U \cap X$ as open in $E$ if $U$ is open in $X$ .

Let $X$ be a topological space. A connected subset of $X$ is a subset $E \subseteq X$ such that the subspace $E$ of $X$ is a connected topological space.

Compact Sets

Let $X$ be a set. An untrafilter on $X$ is a filter $ℱ$ such that there is no filter on $X$ which is strichly finer than $ℱ$ .

Let $X$ be a topological space. The space $X$ is quasicompact if every filter on $X$ has a cluster point.

A topological space is compact if it is quasicompact and Hausdorff.

Let $X$ be a set. A filter $ℱ$ on $X$ is convergent if it has a limit point.

Let $X$ be a topological space. The following are equivalent.

Every filter on $X$ has at least one cluster point.
Every ultrafilter on $X$ is convergent.
Every family of closed subsets of $X$ whose intersection is empty contains a finite subfamily whose intersection is empty.
Every open cover [IF NECESSARY, HAS 'COVER' BEEN DEFINED] of $X$ contains a finite subcover.

Proof.

(b) $\Rightarrow$ (a):
Let $ℱ$ be a filter and let $\overline{ℱ}$ be an ultrafilter containing $ℱ$ . Let $x$ be a limit point of $\overline{ℱ}$ . Then $x$ is a limit point of $ℱ$ .
(a) $\Rightarrow$ (b):
Let $\overline{ℱ}$ be an ultrafilter. Let $x$ be a clusterpoint of $\overline{ℱ}$ . Then $x$ is a limit point of $\overline{ℱ}$ . So $\overline{ℱ}$ is convergent.
(a) $\Rightarrow$ (c):
Let $𝒞$ be a closed family with empty intersection. If every finite subfamily has empty intersection then $𝒞$ generates a filter $ℱ$ . Let $x$ be a cluster point of $ℱ$ . Then $x \in C$ for all $C$ in $𝒞$ . This is a contradiction. So there exists a finite subfamily that does not have empty intersection.
(c) $\Rightarrow$ (a):
If there exists a filter $ℱ$ without a cluster point then $𝒞 = \{\overline{F} | f \in ℱ\}$ [???] IS THIS CORRECT? is a family of closed sets contradicting (c).
(c) $\Leftrightarrow$ (d):
By taking complements.

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Let $X$ be a metric space and let $E$ be a subset of $X$ . The set $E$ is compact if and only if every infinite subset of $E$ has a limit point in $E$ .

Proof.

$\Leftarrow$ :
Let $K$ be a compact set and let $E$ be an infinite subset of $K$ . If there is no limit point of $E$ in $K$ then for each $p \in K$ there is a neighbourhood $N_{p}$ which contains no other element of $E$ . Then the open cover $𝒩 = \{N_{p} | p \in K\},$ of $K$ has no finite subcover.
$\Rightarrow$ :
Let $S$ be an infinite subset of $E$ . The metric space $E$ has a countable base. So every open cover of $E$ has a countable subcover $𝒞 = \{C_{1}, C_{2}, \dots\}$ . If $𝒞$ does not have a finite subcover, then for each $n$ , ${(C_{\leq n})}^{c} \neq \emptyset but \underset{n}{\cap} C_{\leq n}^{c} = \emptyset .$ Let $S$ be a set which contains a point from each $C_{\leq n}^{c}$ . Then $S$ has a limit point. But this is a contradiction.

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Let $X$ be a Hausdorff topological space and let $K$ be a compact subset of $X$ . Then $K$ is closed.

Proof.

Let $x \in \overline{K}$ . The neighbourhood filter $ℬ (x)$ of $x$ induces a filter $ℬ_{K}$ on $K$ which has a cluster point $y \in K$ . Since $ℬ (x)$ is coarser than $ℬ_{K}$ (considered as a filter base on $X$ ) the point $y$ is a cluster point of $ℬ (x)$ . So $y = x$ since $X$ is Hausdorff.

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The proof in Baby Rudin [R]:
We will show that $K^{c}$ is open. Let $p \in K^{c}$ . Let $𝒩$ be the open cover of $K$ given by $𝒩 = \{N_{q} | q \in K\}, where N_{q} = \{B_{\frac{1}{2} d (p, q)} (q) | q \in K\} .$ Let $\{N_{q_{1}}, \dots, N_{q_{l}}\}$ be a finite subcover of $K$ . Then $M = M_{q_{1}}∩ \dots∩ M_{q_{l}}, where M_{q} = B_{\frac{1}{2} d (p, q)} (p),$ is an open set such that $p \in M \subseteq K^{c}$ . So $p$ is an interioe point of $K^{c}$ . So $K$ is open.

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Let $X$ be a metric space and let $E$ be a compact subset of $X$ . Then $E$ is closed and bounded.

Proof.

Since a metric space is Hausdorff, $E$ is closed. If $E$ is not bounded then there is an infinite sequence in $E$ that does not have a limit point.

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A k-cell is compact. [DEFINE K-CELL?]
Let $E$ be a subset of $ℝ^{k}$ . If $E$ is closed and bounded, then $E$ is compact.

Proof.

If $E$ is closed and bounded then $E$ is a closed subset of a $k$ -cell. Since closed subsets of compact sets are compact $E$ is compact.

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References

[R] W. Rudin, Principles of Mathematical Analysis (3rd ed.), McGraw-Hill , (1976).