# Adherent Point, Accumulation Point, Boundary Point, Interior Point

I'm drawing certain cases to better understand some of the jargon in introductory Topology. Can you guys quickly tell me if the four statements I made below are correct? I'm doubtful about the second statement the most.

Also is notion of accumulation points and adherent points generalizable to all topological spaces or like the definition states does it only hold in a Euclidean space? Thanks!

All definitions are relative to the space in which S is either open or closed below. These definitions are from Rudin's Principles of Math. Analysis

and these are from Apostol's Mathematical Analysis

Are the following statements true

1. In 2, if we take $S$ to be closed relative to $X$, then $x$ would be an interior point of S

2. In 2, this is the only case where $x$ is an adherent point but not an accumulation point

3. In 3, if we take the entirety of the neighborhood $N_{r}(x)$ and not just the shaded part then $x$ is not longer an interior point

4. The criteria for a boundary point is that any neighborhood of $x$ has nonempty intersection with either $S$ or $S^c$

• For 2, take an open neighborhood around $x$, remove $x$ from this neighborhood, and then ask yourself whether there are any other points of your set (the blue one I assume) in your neighborhood. – rubikscube09 Oct 15 '17 at 19:23
• @rubikscube09 Right, and from you say we can make $S$ closed, making $S^c$ open then $x$ is not an accumulation point of $S^c$ but still an adherent point? Thanks! – Red Oct 15 '17 at 19:42
• I'm having a bit of issue understanding your labels. What color is set $S$ and its complement, and which set are we working with? I.e do we want to see if $x$ is an adherent point/accum point of $S$ or $S^c$? – rubikscube09 Oct 15 '17 at 20:15
• @rubikscube09 yeah sorry about that, the biigger circle is the set $S$ and the smaller circle is a neighborhood of $x$, blue just indicates the space or universe in which $S$ is open or $S^c$ is closed and red is the overlap between the universe and a neighborhood. – Red Oct 15 '17 at 20:44
• Ok, I've typed a full response. – rubikscube09 Oct 15 '17 at 21:07

1. Yes, $x$ is an interior point (of $S$). It is irrelevant if $S$ is open or closed, what you showed was that we could draw an open neighborhood around $x$ that was totally contained in $S$.
2. Based on the drawing, it seems that $x$ is an accumulation point of $S$, as for any open neighborhood you take around it (and remove $x$ from) contains a point of $S$. This is because it is on the boundary, so every open set around it contains some point in the set and outside it. Note that every accumulation point of aset has to be an adherent point(why?).
3. I think you may have this backwards. In its current state, $x$ is not an interior point (of the shaded area) but if you shade in the rest of the neighborhood, it will become one. This is because the way it is drawn, any open ball around $x$ will contain some point outside the shaded region and thus it cannot be an interior point.