# Is $X = [0,1] \bigcup [2,3]$ connected?

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 $$[0,1]$$, $$[2,3]$$.
• Why should "clearly not path connected" be a reason for connectednesss? – Hagen von Eitzen May 26 '19 at 15:31
• Because If X is complete then path connected implies connectedness and path disconnected implies disconnected? – Qwertford May 26 '19 at 15:42

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.

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.

"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 $$\Bbb R$$.

• @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? – Henno Brandsma May 26 '19 at 15:44
• 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. – Qwertford May 26 '19 at 15:45
• You need to understand the subspace topology. – Randall May 26 '19 at 17:13