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in $\mathbb{R}$, with the usual topology, is the set $A = \mathbb{[a,b]}$ connected? what about $B = [a,b] \cup [c,d]$ where $a < b < c < d$, i would say $A$ is connected while $B$ is not.

However i cannot understand how to apply the definition to prove/disprove they're connected.

Munkres says

Let $X$ be a topological space. A separation of $X$ is a pair $U,V$ of disjoint nonempty open subsets of $X$ whose union is in $X$. The space $X$ is said to be connected if there not exist a separation of $X$.

So to prove that $A$ is connected i have to prove that there's no separation (i would try by contradiction), while to prove $B$ is not connected i would try to find such pair of open subsets... however i'm a bit confused on how to apply the definition since my sets are actual union of closed set.

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You must remember that the set $B$ is a topological space with induced topology*. The subspace $B$ isn't connected because $B\cap (-\infty,f)\cup(f,\infty)\cap B$ with $b<f<c$ is a separetion of $B$.

*If $X$ is a topological space and $Y\subseteq X$, then $Y$ is a topological space with the following family of open sets : $Z\subseteq Y$ is open if and only if exist an open set $A\subseteq X$ such that $A\cap Y=Z$ .

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  • $\begingroup$ Could you elaborate more? $\endgroup$ – user8469759 Jan 25 '16 at 16:08
  • $\begingroup$ Yes, I have edited my answer :) $\endgroup$ – Vincenzo Zaccaro Jan 25 '16 at 16:09
  • $\begingroup$ So you're saying that i should look at $B$ as a subspace of a topological space and apply on such space the definition? $\endgroup$ – user8469759 Jan 25 '16 at 16:12
  • $\begingroup$ Yes :) If $Y\subset X$ you say that $Y$ is connected or sconnected with the subspace topology. $\endgroup$ – Vincenzo Zaccaro Jan 25 '16 at 16:17
  • $\begingroup$ Ok, that's a first step. Now... why $B \cap (-\infty,f)$ and $(f,+\infty) \cap B$ is a separation? why they're open in relation the definition of subspace? $\endgroup$ – user8469759 Jan 25 '16 at 16:19
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The proof that an interval $[a,b]$ is connected is standard but not trivial (i.e., it doesn't follow immediately from the definition). You can use this question as a reference; for completeness, here is the argument invoked there:

Let $Y$ be an interval in $\mathbb{R}$ and suppose that $Y$ is not connected.

Then $Y=A\cup B$, where $A,B\subseteq Y$ are open in $Y$, $A,B\neq\emptyset$ and $A\cap B=\emptyset$. Let $a\in A$ and $b\in B$. Without loss of generality, $a<b$.

Let $\alpha=\sup\{x\in\mathbb{R}:[a,x)\cap Y\subseteq A\}$.

Then $\alpha\le b$ and $\alpha\in Y$. It is clear that $\alpha\in Cl_{Y}(A)$, and $A$ is closed in $Y$, then $\alpha\in A$. Since $A$ is open in $Y$, $Y$ is an interval, $b\in Y\setminus A$ and $\alpha <b$, then there exists $r>0$ such that $(\alpha-r,\alpha+r)\cap Y\subseteq A$. We can conclude that $[a,\alpha+r)\cap Y\subseteq A$, which is a contradiction.


To see that $[a,b] \cup [c,d]$ is not connected, note that $[a,b]$ and $[c,d]$ are two open subsets (in the topology of $[a,b] \cup [c,d]$) that partition $[a,b] \cup [c,d]$.

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