$x$ is an adherent point of $X$ iff $\exists (a_n)_{n=0}^\infty$ such that $a_n \in X$, which converges to $x$.

Question

Lemma

Let $X$ be a subset of $\Bbb R$, and let $x\in\Bbb R$. Then $x$ is an adherent point of $X$ if and only if there exists a sequence $(a_n)_{n=0}^\infty$ consisting entirely of elements in $X$, which converges to $x$.

proof:

Since $x$ is an adherent point of $X$, then $\exists y\in X ~s.t.~ |x-y|\leq \epsilon~\forall \epsilon>0$

therefore, you can construct a constant sequence $(a_n)_{n=0}^\infty=y$ that satisfies the conditions namely $y\in X$ and $|a_n-x|\leq \epsilon ~~\forall \epsilon $

since its an iff statement, we need to prove the other direction.

suppose there exists a sequence with the previously mentioned sequence conditions, then for any choice of $\epsilon$ there exists an element $a_n$ such that $|a_n-x|\leq\epsilon$

therefore, $x$ is an adherent point.

Is my proof correct? The book mentions that I should use axiom of choice in one of the implication. but I cant see at what point of the proof the need of that axiom arises?

Answer 1

Let me write a solution for $(\Rightarrow)$: (I hope I am correct in how the AC is used)

$(\Rightarrow)$ Assume that $x$ is an adherent point of $X$. Then for any $n\in \mathbb N$, let $\epsilon =1/n$. Then by definition, the set

$$X_n = \{y \in X: |y-x|<1/n\}$$

is nonempty. So by axiom of choice, we can find $a_n \in X_n \subset X$. Then the sequence $(a_n)_{n=1}^\infty$ in $X$ satisfies

$$|a_n - x| < 1/n$$ for any $X$. One can then show that $\lim_{n\to \infty} a_n = x$. (The use of Axiom of choice is so natural that I don't even aware I am using it.)

Answer 2

You're mixing the quantifiers in the definition of an adherence point. If $x$ is an adherent point of $X$, then for every $\varepsilon>0$ there exists $y\in X$ such that $|x-y|<\varepsilon$.

Otherwise $0$ is not an adherence point of $(0,1)$, but of course there is a sequence converging to it from the interval.

And this is why you haven't used the axiom of choice either. You need the axiom of choice to choose $x_n\in X$ such that $|x-x_n|<\frac1n$, or some other sequence of $\varepsilon$'s. Your mistake allowed you to conclude that you can always take a constant sequence, in which case the axiom of choice isn't needed at all.

Without the axiom of choice it is consistent that there is a dense subset of $\Bbb R$ which has no countably infinite sequence. If $X$ is such set, then every convergent sequence of elements of $X$ is eventually constant, but since $X$ is dense, every element of $\Bbb R$ is an adherence point of $X$. So the use of the axiom of choice is indeed necessary.