Why is the closed interval [0,1] compact under the limit-point definition?

Conditions

• E is the closed interval [0,1].
• Compactness is judged by Hausdorff's limit-point definition.
• S is an infinite subset of [0,1].

Reasoning, step by step

1. Consider an infinite subset S of [0,1], such as {$1/n$ : $n=1$,2,3,...}.
2. Identify the limit point of S, which is 0.
3. Check if the limit point 0 belongs to the set [0,1].
4. Since $0 \in [0,1]$, the condition is satisfied for this subset.
5. The video generalizes this to state that all infinite subsets of [0,1] have limit points in [0,1], making it compact.

Example

The video shows green dots representing {$1/n$} clustering toward 0 on the interval [0,1], with square brackets indicating that 0 is included, and labels it 'Compact $E=[0,1]$'.

Common misconceptions

• Thinking that removing a single endpoint from an interval leaves compactness unchanged.
• Confusing the limit point of a sequence with the limit of the sequence itself without considering the set's closure.

Watch the explanation

Explore next

• Why does removing the endpoint 0 make (0,1] noncompact?
• Are Fréchet compactness and limit-point compactness equivalent in metric spaces?