limit points of uncountable subsets of R^n


PropositionPlanetmathPlanetmath. Let ℝn be an n-dimensional, real normed space and let A⊆ℝn. If A is uncountable, then there exists limit pointPlanetmathPlanetmath of A in ℝn.

Proof. For any k∈ℕ let

𝔹k={v∈ℝn|||v||≤k},

i.e. 𝔹k is a closed ballPlanetmathPlanetmath centered in 0 with radius k. Assume, that for any k the set

Vk=𝔹k∩A

is finite. Then ⋃Vk=A would be at most countableMathworldPlanetmath. ContradictionMathworldPlanetmathPlanetmath, since A is uncountable. Thus, there exists k0∈ℕ such that Vk0 is infiniteMathworldPlanetmath. But Vk0⊆𝔹k0 and since 𝔹k0 is compactPlanetmathPlanetmath (and Vk0 is infinite), then there exists limit point of Vk0 in ℝn. This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof. □

Corollary. If A⊆ℝn is uncountable, then there exist infinitely many limit points of A in ℝn.

Proof. Assume, that there are finitely many limit points of A, namely x1,…,xk∈ℝn. For ε>0 define

Aε={v∈ℝn|∀i||v-xi||>ε}.

Briefly speaking, Aε is a complementPlanetmathPlanetmath of a union of closed balls centered at xi with radii ε. Of course Aε≠∅ since there are finitely many limit points. Let

Vε=A∩Aε.

Assume, that Vε is countable for every ε. Then

A⊆⋃n∈ℕV1n∪{x1,…,xk}

would be at most countable (of course under assumptionPlanetmathPlanetmath of Axiom of ChoiceMathworldPlanetmath). Contradiction. Thus, there is γ>0 such that Vγ is uncountable. Then (due to proposition) there is a limit point x′∈ℝn of Vγ. Note, that

x′∈Vγ¯⊆Vγ′

for some 0<γ′<γ. Thus x′ is different from any xi. Contradiction, since x′ is also a limit point of A. □

Title limit points of uncountable subsets of R^n
Canonical name LimitPointsOfUncountableSubsetsOfRn
Date of creation 2013-03-22 19:07:57
Last modified on 2013-03-22 19:07:57
Owner joking (16130)
Last modified by joking (16130)
Numerical id 6
Author joking (16130)
Entry type Theorem
Classification msc 54A99