Clarification on Boundary and Limit Points in Topology

Question

I'm trying to better understand the concepts of boundary points and limit points in topology. According to the definitions from Wikipedia:

• A point x in a topological space X is a limit point of a subset S if every neighborhood of x contains at least one point of S different from x itself.

• The boundary of a subset S in a topological space X is the set of points p of X such that every neighborhood of p contains at least one point of S and at least one point not of S.

Given these definitions, it seems clear that the two concepts are distinct. However, I've come across a case that seems to contradict the definition of a boundary point. Specifically, when considering the set S = {0} in the real numbers (R) with the usual topology, it appears that 0 does not qualify as a boundary point because every neighborhood of 0 contains only 0, which is a point from S, with no points from R\ S.

Can someone confirm if my understanding is correct, and if not, why is 0 considered a boundary point in this case? Are there any other counterintuitive examples or cases that help illustrate the distinction between boundary and limit points in topology? I'd appreciate a clear explanation to help solidify my understanding of these concepts.

Thank you for your insights on this matter.

Answer 1

From what you say it seems to me you're switching the quantifiers in the definition of a boundary point. To clarify the definition, I'll rewrite it a bit.

Let $X$ be a topological space and $S\subseteq X$. We say that $x\in X$ is a boundary point of $S$, if for any neighborhood $U$ of $x$, there are points $y \in S$ and $z \in X\setminus S$ such that $y,z \in U$. Alternatively, we may say that any neighborhood $U$ of $x$ must intersect both $S$ and $X \setminus S$, ie. $U\cap S \neq \emptyset \neq U \cap (X\setminus S)$.

We don't ask for existence of points $y \in S$ and $z \in X\setminus S$ such that for any neighborhood $U$ of $x$ we have $y,z \in U$ (that, in fact, would imply $y=z$ in "most" spaces).

In your example, ie. $X=\mathbb R$ with the standard topology and $S= \{0\}$, consider any neighborhood $U$ of $0$. Then $U$ is a superset of an open interval $(-\varepsilon,\varepsilon)$ for some $\varepsilon > 0$. Thus $U$ contains both $0 \in S=\{0\}$ and $\frac\varepsilon2 \in X\setminus S = \mathbb R \setminus\{0\}$.