I know that if a topological space $X$ is connected, then the only subsets of $X$ that are both open and closed are $\varnothing$ and $X$.

However, if I take for example the interval $I = [0,1]$ with the discrete topology, then $[0,1/2)$ is open so that the complement $[1/2,1]$ is closed in $I$. However, since we are working with the discrete topology, $[1/2,1]$ is open in $I$. However, since $[0,1]$ is connected, the only sets that are both open and closed are $[0,1]$ and $\varnothing$.

Could anyone point out the flaw in my thoughts?

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    $\begingroup$ $[0,1]$ is connected in the standard topology. In the discrete topology, no space with more than one point is connected. $\endgroup$ – Daniel Fischer Jun 10 '14 at 19:55
  • $\begingroup$ Ah, I see. Thanks! $\endgroup$ – user59999 Jun 10 '14 at 19:56

The flaw is mixing different topologies. Connectedness is a property of the topology, not of the underlying set. $[0,1]$ is connected in the standard topology (and in all coarser topologies), but that doesn't imply that it is connected in the discrete topology. Indeed, a space with the discrete topology is connected only if it is empty or contains exactly one point. Discrete spaces are always totally disconnected (the connected components are singletons).


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