# $p\in E\subset R^n$: limit and/or isolated and/or interior point of $E$?

According to Rudin's Principles of Mathematical Analysis:

• A point $p$ is a limit point of the set $E$ if every neighbourhood of $p$ contains a point $q\ne p$ such that $q\in E$

• If $p\in E$ and $p$ is not a limit point of $E$, then $p$ is called an isolated point of $E$

• A point $p$ is an interior point of $E$ if there is a neighbourhood $N$ of $p$ such that $N\subset E$

I)

Consider $R^3$ with the usual distance metric. Off course, a limit point might not be contained in $E$. But, consider the case where $p\in E$. This means that:

• all neighbourhoods of $p$ will contain some $q$ that is also in $E$
• in fact there will be a neighbourhood of $p$ that is a subset of $E$

So, $p$ it is both a limit point and an interior point of $E$. In consequence, $E$ has no isolated points?

II)

Could you give me some examples of spaces where one can identify points that are only limit, isolated and interior.

• The first bullet under I) is correct, but it is not guaranteed that $p\neq q$, so we cannot conclude yet that $p$ is a limit point of $E$. The second bullet does not have to be true. For both remarks: have a look at set $E=\{p\}$. – drhab Sep 9 '16 at 8:22

Your reasoning is incorrect. Rudin doesn't specifically talk about it as far as I recall, but there exists a boundary of a set, say, $E$, which is defined as those points $p$ such that every neighborhood of $p$ contains a point $q_1 \in E$ and a point $q_2 \in E^c$. In other words, $p$ is a limit point for both $E$ and $E^c$,and hence no neighborhood of $p$ is a subset of $E$.
Consider the set $A \subset \mathbb R$ defined by $A = \{(0,1) \cup {2}\}.$ Then 0 and 1 are both limit points but not interior points, every $x \in (0,1)$ is an interior point, and 2 is an isolated point.
• Thank you. I was referring to points that are not on the boundary and are members of $E$. So, in your example, 1/2 is both a limit point and an interior point, right? – hyperio Sep 9 '16 at 7:32
• If every neighborhood of $p$ has non-empty intersection with $E$ and $E^c$ then $p$ is an element of the boundary of $E$ (wich is the same as the boundary of $E^c$). However, $p$ is not necessarily a limit point of $E$ and $E^c$. Take for instance the set $E=\{p\}$. – drhab Sep 9 '16 at 8:17
• @drhab good point. In this case $p$ would be a limit point of $E^c$ only, as $E$ has no limit points. I didn't think of that. – David Bowman Sep 9 '16 at 8:21