Countably Compact First-Countable Space is Sequentially Compact
Jump to navigation
Jump to search
Theorem
A countably compact first-countable topological space is also sequentially compact.
Proof 1
Follows directly from:
- Infinite Sequence in Countably Compact Space has Accumulation Point
- Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit
$\blacksquare$
Proof 2
Let $T = \struct {S, \tau}$ be a countably compact first-countable topological space.
By definition of sequentially compact, it is sufficient to show that every infinite sequence in $S$ has a convergent subsequence.
Let $\sequence {s_n}$ be any sequence in $S$.
By Infinite Sequence in Countably Compact Space has Accumulation Point, $\sequence {s_n}$ has an accumulation point $p \in S$.
As $T$ is first-countable, $p$ has a countable local basis, say:
- $\set {V_n: V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots}$
Then a subsequence $\sequence {s_{n_i} }$, where $s_{n_i} \in V_i$, converges to $p$.
This article contains statements that are justified by handwavery. In particular: This last statement is "by intimidation"; justification needed You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding precise reasons why such statements hold. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Handwaving}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
$\blacksquare$