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Let $X = [0,1] \cup [2,3]$ be a metric space with the euclidean metric, is it connected?

Im not sure because i think..

  1. It is connected because $X$ is complete and clearly not path connected.
  2. It is incomplete because $X$ cannot be written as the union of 2 non empty disjoint open sets. They can only be written as the union of 2 non empty disjoint closed sets $[0,1]$, $[2,3]$.
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  • $\begingroup$ Why should "clearly not path connected" be a reason for connectednesss? $\endgroup$ – Hagen von Eitzen May 26 '19 at 15:31
  • $\begingroup$ Because If X is complete then path connected implies connectedness and path disconnected implies disconnected? $\endgroup$ – Qwertford May 26 '19 at 15:42
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Recall that $A \subset X$ is connected if there is a separation of $A$: a set $C \subset A$ which is clopen in $A$.

Now, set $A = [0,1] \cup [2,3]$.

$[0,1] = A \cap [0,1]$ so it is closed in $A$. Similarly $[2,3]$ is closed in $A$.

Observe that $A - [0,1] = [2,3]$ which is open in $A$ because $A \cap (1.5,3.5) = [2,3]$. Hence $[0,1]$ is clopen in $A$, and $A$ is therefore not connected.

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Define $f(x)$ to be 0 for $x\in[0,1]$ and $1$ for $x\in[2,3]$. This function is continuous on the specfied domain. Your answer is right there.

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"Disjoint open sets" refers to the topology of $X$, not to some topology of an embedding space (for which you presumably take $\Bbb R$). A subset of the space $X$ is understood to be open when it is so in the topology inherited from $\Bbb R$, aka. "relatively open". That is, $A\subseteq X$ is open in $X$ iff there exists an open subset of $\Bbb R$ such that $A=X\cap U$. Now note that $(-1,2)$ and $(1,4)$ are open subsets of $\Bbb R$.

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  • $\begingroup$ @Qwertford also note that $[0,1]$ and $[2,3]$ are each other's complement in $X$, so if one is closed, the other is open and vice versa. You admitted they were closed, right? $\endgroup$ – Henno Brandsma May 26 '19 at 15:44
  • $\begingroup$ What about the defintion of open as A is open if for every elemtent a of A, the neighbourhood around a is contained in A? Then [0,1] is not opened because it contains 0 and 1 with neighbours which are not subsets of A. $\endgroup$ – Qwertford May 26 '19 at 15:45
  • $\begingroup$ You need to understand the subspace topology. $\endgroup$ – Randall May 26 '19 at 17:13

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