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I'm starting to learn topology watching "The WE-Heraeus International Winter School on Gravity and Light" series on YouTube in which it uses open sets and neighborhood to define Topology. In the neighborhood definition, it defines a soft ball and claims that for open set $O$, $U \in O$ means for all $p \in U$ there exists

$r \in R^+: B_r(p) \subseteq U$, where $ B_r := \{(q_1, ...q_d)| \sum_{n=1}^{d} {(q_1 - p_i)}^{2} < r \}$

where $B_r(p)$ is the soft ball. However to me there seems to be an inconsistency in two definitions because aside from the empty set, $U$ is comprised of sets of rank 1 or greater than 1 where sets of rank greater than 1 are the neighborhoods. But according to the open set definition, any arbitrary union of subsets is also a subset of $U$. So if we have points or rank 1 subsets $\{x\}$ and $\{y\}$ in $U$ which are far apart, $\{x, y\}$ must also be a subset. But $\{x, y\}$ fails to satisfy the neighborhood definition since it does not describe a neighborhood but are two points that is just far apart. So have my thinking gone wrong?

Martin Sleziak
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I feel some type errors here. I get that $O$ (often writen caligraphically as $\mathcal{O}$) is the collection of all open sets (not an open set as you write at one place), and that $U$ is an open set – it is a set of points, so it is not comprised of sets (unless you mean that its a union of sets).

If $\{x\}$ is a neighborhood of $x$, then $\{x\}$ is itself open. Such point is called isolated. There is no contradition in an open set $U$ containing two isolated points $x$ and $y$ – it is true that $\{x, y\} ⊆ U$ is and open set and a neighborhood of both $x$ and $y$. Being far apart is irrelevant here.

Also it seems that you really work in $ℝ^n$. There no point is isolated, so $\{x\}$ is never a neighborhood of $x$.

Adam Bartoš
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