True or false: in a compact topological space every infinite set has a limit point. Justify your answer.
Solution
The statement is true.
1. Let \(A\) be an infinite subset of a compact space \(X\) and suppose, for contradiction, that \(A\) has no limit point.Then for every \(x\in X\) there is an open neighborhood \(U_x\) of \(x\) such that\[ U_x\cap(A\setminus\{x\})=\varnothing. \]
A point \(x\in X\) is a limit point of \(A\) if every open neighborhood of \(x\) meets \[ A\setminus\{x\}. \] If \(x\) is not a limit point, the negation of this condition gives an open neighborhood \(U_x\) with \[ U_x\cap(A\setminus\{x\})=\varnothing. \] By the assumption that \(A\) has no limit point, this holds for every \(x\in X\).
2. The family\[ \{U_x:x\in X\} \]is an open cover of \(X\), so finitely many members\[ U_{x_1},\ldots,U_{x_n} \]cover \(X\).
Each \(x\) belongs to its own neighborhood \(U_x\), so the family covers \(X\). Compactness gives a finite subcover.
3. The finite subcover forces\[ A\subseteq\{x_1,\ldots,x_n\}. \]
Take \(a\in A\). Since the chosen neighborhoods cover \(X\), there is an \(i\) with \[ a\in U_{x_i}. \] By 1, \[ U_{x_i}\cap(A\setminus\{x_i\})=\varnothing. \] Because \(a\in A\cap U_{x_i}\), it follows that \[ a=x_i. \] Thus every point of \(A\) lies among \(x_1,\ldots,x_n\).
4. Therefore every infinite subset of a compact topological space has a limit point.
By 3, the assumption in 1 would make \(A\) a subset of a finite set, hence finite, contradicting the hypothesis that \(A\) is infinite. Therefore \(A\) must have a limit point.