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Any set of the form $\{x, y\}$ is disconnected. Wouldn't this imply that the rational numbers is a discrete space, since $\{x\}$ and $\{y\}$ are open?

Math1000
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Mike
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1 Answers1

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An open set of the subspace topology on $\Bbb Q$ is obtained by taking the intersection of $\Bbb Q$ with an open set of $\Bbb R$. Any non-empty open set in $\Bbb R$ contains an interval. So, could $\{x\}$ possibly be open in $\Bbb Q$?

David Mitra
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  • Isn't {x} open in the subspace topology on {x, y}? – Mike Apr 27 '12 at 21:04
  • Yes, but is ${x,y}$ open in $\mathbb Q$? – Gunnar Þór Magnússon Apr 27 '12 at 21:08
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    @Mike ${x}$ is open in the space ${x,y}$. But singleton sets are not open in $\Bbb Q$. This is so because you cannot write ${x}=O\cap \Bbb Q$ with $O$ an open set of $\Bbb R$ (a non-empty open subset of $\Bbb R$ contains infinitely many rationals). – David Mitra Apr 27 '12 at 21:18
  • Okay thanks. I forgot that a subset of a space being connected/disconnected means being connected/disconnected in its subspace topology. – Mike Apr 27 '12 at 21:21
  • @Mike ${x}$ is open in ${x}$ but that doesn't mean that ${x}$ is open in $\mathbb R$ or that $\mathbb R$ is discrete. – bof Jul 23 '16 at 11:04