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We say $X$ is disconnected if it can be written as the union of two disjoint nonempty open subsets of $X$. If $ Y $ is a subspace of $X$, when we say $Y$ is disconnected, do we mean with respect to the topology on $X$, or the subspace topology induced on $Y$? Can we say $Y$ is disconnected if it is contained in the union of two disjoint nonempty open subsets of $X$?

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2 Answers 2

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We cannot say that $Y$ is disconnected if it is contained in the union of two disjoint nonempty open subsets of $X$. For example, if $X=\mathbb{R}$ and $Y=(1,2)$, then $$Y\subset (-1,0)\cup (1,2)$$ even though $Y$ is connected.

If $X$ is a topological space and $Y\subseteq X$, then certainly, if $Y$ is disconnected as a subset of $X$, i.e. if $Y=A\cup B$ for non-empty disjoint open $A,B\subset X$, then it is disconnected in its subspace topology, because both $A$ and $B$ are open subsets of $Y$ in the subspace topology.

However, the opposite direction is not true. That is, if $Y=U\cup V$ for non-empty disjoint open $U,V\subset Y$, then we have that $U=Y\cap A$ and $V=Y\cap B$ for some non-empty open $A,B\subseteq X$, but it need not be the case that we can always find $A$ and $B$ that are disjoint.

For example, let $X=\{1,2,3\}$ given the topology $T=\{\varnothing,\{2\},\{1,2\},\{2,3\},X\}$. Let $Y=\{1,3\}\subset X$. Then $Y$ is disconnected in the subspace topology, as $U=\{1\}$ and $V=\{3\}$ are disjoint non-empty sets that are open in the subspace topology of $Y$ and whose union is $Y$, but $Y$ cannot be covered by disjoint non-empty open subsets of $X$.

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Can $Y$ be a subspace if it only consists of the point $(1,2)$? –  Théophile Mar 15 '12 at 0:52
    
$(1,2)$ is not a point, it is the open interval $(1,2)$. –  Zev Chonoles Mar 15 '12 at 0:54
    
You can’t guarantee that $A\cap B=\varnothing$; see my addendum to your answer. –  Brian M. Scott Mar 15 '12 at 1:00
    
@Brian: Yup, it just took me a bit to come up with an example :) +1 to your answer. –  Zev Chonoles Mar 15 '12 at 1:01

To add to Zev’s answer, we can’t even say that $Y$ is disconnected if and only if it is contained in the union of two disjoint open subsets of $X$ that both have non-empty intersection with $Y$. To see this, let $X=\Bbb R$, and declare $U\subseteq X$ to be open iff $U=\varnothing$ or $0\in U$; it’s easy to check that this really is a topology on $X$. Let $Y=X\setminus\{0\}$. The relative topology on $Y$ is discrete, so $Y$ is very badly disconnected. However, all non-empty open sets in $X$ contain $0$, so no two of them are disjoint, and therefore it’s impossible to find open $U,V\subseteq X$ such that $Y\subseteq U\cup V$ and $U\cap V=\varnothing$.

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