Let $(E,\tau)$ be a topological space, $\mathcal N_\tau(x):=\{N\subseteq E:N\text{ is a }\tau\text{-neighborhood of }B\}$ for $x\in E$ and $B\subseteq E$. Remember that $x\in E$ is called
- $\tau$-accumulation point of $B$ if $$\forall N\in\mathcal N_\tau(x):B\cap N\setminus\{x\}\ne\emptyset;\tag1$$
- $\tau$-isolated point of $B$ if $$\exists N\in\mathcal N_\tau(x):B\cap N=\{x\}\tag2.$$
Moreover, $B$ is called $\tau$-discrete if every point in $B$ is an $\tau$-isolated point of $B$; i.e. $$\forall x\in B:\exists N\in\mathcal N_\tau(x):B\cap N=\{x\}\tag3.$$
Assume $B$ is $\tau$-discrete. Can we show that $B$ has no $\tau$-accumulation point or do we need to additionally assume that $B$ is $\tau$-closed (i.e. $E\setminus B\in\tau$)?
Let $x\in E$. If $x\in B$, then $$B\cap N=\{x\}\tag4$$ and hence $$B\cap N\setminus\{x\}=\emptyset\tag5$$ for some $N\in\mathcal N_\tau(x)$; hence $x$ is not a $\tau$-accumulation point of $B$.
Now assume $x\in E\setminus B$. If $B$ would be $\tau$-closed, we would clearly find a $N\in\mathcal N_\tau(x)$ with $N\subseteq E\setminus B$ and hence $(5)$ would hold again and we we could conclude that $x$ is not a $\tau$-accumulation point of $B$.
But do we really need the $\tau$-closedness assumption? If so, can it be shown that the following equivalence holds: $B$ is $\tau$-closed and $\tau$-discrete if and only if $B$ has no $\tau$-accumulation point?
Best Answer
A set $B$ being $\tau$-discrete does not imply it has no accumulation points, as the standard example $B=\{\frac1n\mid n \in \Bbb N^+\}$ in $\Bbb R$ in the standard topology shows. Here $0 \in B'$.
If $B$ is closed and $\tau$-discrete then it indeed does not have an accumulation point: if $p$ were one, it must lie in $B$ by closedness and this would contradict the discreteness of $B$.
And the condition is necessary of course: a set $B$ has no acculation points iff it is closed and $\tau$-discrete. We just saw the left to right implication, and if a set has no accumulation points, it is closed (voidly, as a set is closed iff it contains all its accumulation points, which is trvial when there are none) and $\tau$-discrete (as no $b \in B$ is an accumulation point it must be an isolated point (it's one or the other).