- #1
Bashyboy
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Homework Statement
Problem: Let ##A## be an infinite subset of a ##T_1## space, and let ##x## be a limit point of ##A##. Prove that every open neighborhood of ##x## contains infinitely many points of ##A##.
Homework Equations
The Attempt at a Solution
First note that if ##\mathcal{O}## is an open neighborhood of ##x## not containing ##y##, then there exists open neighborhoods ##U## and ##V## containing ##x## and ##y##, respectively, but not containing the other point. Hence, ##\mathcal{O} \cap U \subseteq \mathcal{O}## is an open set containing ##x## but not containing ##y##.
Proof of main theorem: Let ##\mathcal{O}## be an open neighborhood of ##x##. Then there exists ##a_1 \in \mathcal{O} \cap (A-\{x\})##. By the observation made above, we can find an open neighborhood ##\mathcal{O}_2## of ##x## that doesn't contain ##a_1## and ##\mathcal{O}_1 \subseteq \mathcal{O}##. Hence, there exists ##a_2 \in \mathcal{O}_2 \cap (A-\{x,a_1\})##. Hence ##(a_n)## is an infinite sequence of distinct points in ##A## such that ##a_n## is in ##\mathcal{O}## for every ##n \in \mathbb{N}##, thereby proving that ##\mathcal{O}## contains an infinite number of points in ##A ~ \Box##.How does this sound? It seems little shaky; perhaps it could be made slightly more rigorous.