Why is $(.5, 1]$ considered an open set in $[0, 1]$? This is from a topology textbook.
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2$(.5,1]=(.5,a)\cap [0,1]$ with $a>1$. – Beni Bogosel Oct 01 '11 at 07:04
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1Because its complement is closed :) (The other comments and answers speak to the real answer to your question.) – 2'5 9'2 Oct 01 '11 at 07:16
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Do you know what can be called as an open set in $[0,1]$? – Oct 01 '11 at 14:16
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@alex.jordan I thought about that but wasn't sure if it was true in general. – Jeff Oct 01 '11 at 16:49
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Suppose $X$ is a topological space, and $S\subseteq X$ is any subset of $X$. Then the subspace topology on $S$ consists of $$\{U\subseteq S\mid\exists\text{ an open }V\subseteq X\text{ such that }U=V\cap S\}$$ So take $X=\mathbb{R}$, and $S=[0,1]$. Can you think of an open subset $V\subseteq\mathbb{R}$ such that $V\cap S=(0.5,1]$?
Zev Chonoles
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The open sets on a subspace $ S \subseteq X$ are simply the $ U \cap S,$ where $U$ is any open set of $X.$ In this case, $X$ is the real line, the interval $(1/2,3)$ is an open set in $X,$ and the intersection with $S = [0,1]$ is therefore open in the subspace topology. It is likely to be worth your time to show that this definition does give a topology on a subset.
Will Jagy
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If you think of $[0,1]$ as a metric space, with the usual metric restricted to it, then your set is the open ball (in the metric space $[0,1]$) of radius $0.5$ with center $1$. Open balls are open sets.
GEdgar
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