0
$\begingroup$

I know the definition of an adherent point to a set $S$, a point $a$ is adherent to $S$ is given $\epsilon$ there exists am $x\in S$ such that $|x-a|<\epsilon$.

However I am having trouble understanding the proof for the following statement: if $S$ is a subset of $R$, every adherent point to $S$ is the limit of a sequence in $S$.

In particular it is the first part of the proof that I do not get, it says:

If $v$ is adherent to $S$ then given $n$ we can find $x_n \in S$ such that $|x_n -v|<1/n$. But I do not see how this follows from the fact that $v$ is adherent to $S$.

$\endgroup$
1
$\begingroup$

First of all, the standard term for adherent point is limit point. Now, we suppose that $v$ is a limit (adherent) point of $S$. Let $\epsilon=\frac 1n$, since $v$ is a limit point of $S$ there exist a point $x_n\in S$ such that $|x_n-v|<\epsilon$, therefor $|x_n-v|<\frac 1n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.