Accumulation point of open interval in real numbers

Asked May 14 '20 at 12:57

Active May 14 '20 at 12:57

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In book General Topology by Kelley it is stated:
If $A=(0,1)$ is a subset of topological space $(\Bbb R,\tau)$,
then every point in closed interval $[0,1]$ is an accumulation point of $A$.

I understand that in topological space, accumulation points of $[0,1]\subset \Bbb R$ are $0$ and $1$.
But how is entire closed interval a set of accumulation points of the same open interval?

asked May 14 '20 at 12:57

flowian

every point in $[0,1]$ is an accumulation point of $[0,1]$ in the usual topology on $\mathbb R$ – J. W. Tanner May 14 '20 at 13:01
It doesn't hold for any topology defined on $\mathbb{R}$. For example if a topology is induced by a metric $d(x,y) = |x-y| + |sgn(x) - sgn(y)|$ then $0$ is not an accumulation (limit) point of any set. Did you mean an euclidean topology in your question? – Matthew W May 14 '20 at 13:07
@MatthewW I guess it is usual topology, since the book is introduction to topology for graduates. – flowian May 14 '20 at 13:32
3

Does this answer your question? The limit points of an open interval (open set) – Matthew W May 14 '20 at 14:11
@MatthewW, yes it does. thank you. – flowian May 14 '20 at 14:43

Accumulation point of open interval in real numbers

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