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 $E$.


2 Answers 2


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 $X$.

  • $\begingroup$ Hence for any metric space with a metric other than discrete metric interior points should be limit points. $\endgroup$
    – Madhu
    Jul 25, 2018 at 11:49
  • $\begingroup$ And without isolated points (in the chosen metric) $\endgroup$ Jul 25, 2018 at 12:34
  • $\begingroup$ Is this argument valid for every metric space that is also connected? $\endgroup$ Nov 15 at 9:55

We know that every point in an open set is interior point right? Then, if a interior point is always a limit point, doesn't this imply that an open set is always closed? As open set consists of interior points and if they are limit points which are inside the set.


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