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I am reading a proof of Tietze extension theorem from a book Deimling - Nonlinear functional analysis, page 6.

What is the plain meaning of following sentence:

"Since $A \subset \mathbb{R}^n$ is compact, there exists a dense and at most denumerable subset $\{ a_1,a_2, ... \}$ of $A$"

Those $a_i$'s are points in $A$. I don't like that language. Does it say that for every $x\in A$ there is a sequence $\{ x_1,x_2, ... \} \subset A$ so that $x_i \rightarrow x$, as $i \rightarrow \infty$?

Hulkster
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2 Answers2

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If $A$ is endowed with a metric. For every $n>0$, consider the covering $B(x,1/n),x\in A$. You can extract a finite subset $B(x^n_{i_1},1/n),...,B(x^n_{i_n},1/n)$ which cover $A$. The set of points $(x^n_{i_j})$ is dense and numerable.

Let $x\in A$, $U$ an open subset of $A$ containing $x$. There exists $B(x,c)\subset U$. Let $n$ such that $1/n<c$, $x\in B(x^n_{i_j},1/n)$ this implies that $x^n_{i_j}\in B(x,c)$. done.

Tsemo Aristide
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This means that because we have a compact subset $A$ of $\mathbb{R}^n$, there exists a countable dense subset. Informally, there exists a subset $U\subset A$ such that every element of $A$ is arbitrarily close to some element of $U$. This subset $U$ is dense in $A$.

That is, $\forall\epsilon>0$ given, and for all $a\in A$, $\exists u\in U$ such that $$ d(a,u)<\epsilon.$$ The general (metric-free) definition states that any open neighborhood in $A$ about $a\in A$ contains some point $u\in U$.

Take for example the rationals as a countable dense subset of $\mathbb{R}$.

Alekos Robotis
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  • Thanks for answering. That notion of sequence puzzled me, but of course we can list all rational numbers, so there is a "dense sequence" for every compact set. – Hulkster Feb 12 '16 at 17:17