0
$\begingroup$

According to Rudin a point $p$ is said to be a limit point of a subset $E$ of a metric space $X$ if every neighborhood $N$ of $p$ contains a point $q \neq 0$ such that $q\in E$.

A point a point $p$ is said to be an interior point of a subset $E$ of a metric space $X$ if there exist a neighborhood $N$ of $p$ such that $N \subset E$.

Examples given shows that there may be limit points which are not interior points : $(a,b)$. There may be sets which does not contain any of its limit points and no point of it is an interior point.

Is interior points of a subset $E$ of a metric space $X$ is always a limit point of $E$ ? If $p$ is an interior point of $E$ then there exist a a neighborhood $N$ of radius $r$ such that $N \subset E$. Then any neighborhood of $p$ having radius less than $r$ will be contained in $N$ and hence contained in $E$. Any neighborhood of $p$ having radius greater than $r$ contains $N$ and hence intersect $E$ at some points other than $p$. Hence $p$ is a limit point of $E$. Is my argument is correct ?

Generally - In topology- this proof will not work as it does not have radius concept. But I want a counter example from metric space if I am wrong. Particularly a subset $E$ of $\mathbb{R}$ which has an interior point which is not a limit point of $E$.

$\endgroup$
0
$\begingroup$

If you are asking about $\mathbb{R}$ with the (standard) Euclidean metric, then the answer is yes:

Suppose that $X\subseteq\mathbb{R}$ and that $x\in X$ is an interior point. Then, by the definition of an interior point, there is some open set $U$ so that $x\in U$ and $U\subseteq X$. Since the topology on $\mathbb{R}$ is generated by open intervals, we know that there is an open interval $I$ so that $x\in I$, $I\subseteq U\subseteq X$. Observe that for $n$ sufficiently large, $x+\frac{1}{n}$ is a sequence of points distinct from $x$ and in $I$ (and hence in $X$) that converge to $x$. Therefore, $x$ is a limit point.

The reason that the argument above worked is that there are infinitely many points in any open interval. If, instead, we use a different metric, the answer to your question is no:

Consider the discrete metric on $\mathbb{R}$, i.e., $$ d(x,y)=\begin{cases}0&x=y\\ 1&x\not=y \end{cases}. $$ Suppose that $X\subseteq\mathbb{R}$ and that $x\in X$. We observe that $x$ is an interior point because the open ball $B\left(x,\frac{1}{2}\right)=\{x\}\subseteq X$. On the other hand, $x$ is not a limit point because the open ball $B\left(x,\frac{1}{2}\right)$ is an open set containing $x$ and no other points of $X$.

$\endgroup$
  • $\begingroup$ Hence for any metric space with a metric other than discrete metric interior points should be limit points. $\endgroup$ – Madhu Jul 25 '18 at 11:49
  • $\begingroup$ And without isolated points (in the chosen metric) $\endgroup$ – Michael Burr Jul 25 '18 at 12:34

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.