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Set [1,2] $\cup$ [3,4] is clopen in $\mathbb R$ ?

If yes, then union of two disjoint closed sets always clopen ?

For above example, can we use following analogy ? Here, given set is closed because it is union of closed sets and open because some sequence may have limit points in (2,3).

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  • $\begingroup$ A set is open if and only if it equals its interior. Can you find any points in that set that are not interior points? If so, the set is not open. $\endgroup$ – Jack Apr 9 '17 at 4:01
  • $\begingroup$ Look at the points 1, 2, 3 and 4. Are they interior points or not? $\endgroup$ – Juniven Apr 9 '17 at 4:06
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    $\begingroup$ "Set [1,2] ∪ [3,4] is clopen in R ?" Nope. It's closed. It isn't open. Why did you think it was? "If yes, then union of two disjoint closed sets always clopen ?" No. Why would you think that is true. It's almost never true. For the reals, I can only think of two cases were it is true; when both sets are empty or one is empty and the other is the entire reals. There are no other options. $\endgroup$ – fleablood Apr 9 '17 at 4:52
  • $\begingroup$ @fleablood Please See following question. math.stackexchange.com/questions/2225771/connectedness-question $\endgroup$ – MeetR Apr 9 '17 at 4:59
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The only clopen sets in $\mathbb{R}$ are itself and $\emptyset$. To see why this set isn't open, consider its complement and look at the sequence $(2 + (k + 1)^{-1})_{k \in \mathbb{N}}$, whose limit is $2$. The sequence doesn't converge in the complement of $[1, 2] \cup [3, 4]$, so complement of set isn't closed so given set isn't open.

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There is no open ball of center $3$ and positive radius that can lie in the set under consideration; by definition the set under consideration is not open.

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  • $\begingroup$ What about that generalized statement ? union of two disjoint closed sets always clopen ? $\endgroup$ – MeetR Apr 9 '17 at 4:05
  • $\begingroup$ Are you thinking that if two open sets are disjoint then their union is open?@Meet Rayvadera $\endgroup$ – Juniven Apr 9 '17 at 4:12
  • $\begingroup$ In pugh's analysis, one example is given like this. "The union of two disjoint closed intervals is not homeomorphic to a single interval. One set is disconnected and the other is connected." I am not able to understand this statement. $\endgroup$ – MeetR Apr 9 '17 at 4:18
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    $\begingroup$ "union of two disjoint closed sets always clopen ?" Not in the least bit true. In the reals, the only three times it is true is when $A =\emptyset; B= \emptyset$ or $A = \emptyset; B=\mathbb R$ or $A = \mathbb R; B= \emptyset$. The the reals, no other cases exist. $\endgroup$ – fleablood Apr 9 '17 at 4:49
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    $\begingroup$ The statement has nothing to do with being closed and open. It says the union of two disjoint intervals can never be a connected interval. $[0,3] \cup [4,5] \ne [a,b]$ for any $a$ and $b$. That's all. It has nothing to do with clopen. BTW the union of two closed sets, disjoint or not, is always closed. There is no reason, and it'd be very rare, that they'd be open. $\endgroup$ – fleablood Apr 9 '17 at 4:56
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No, $X = [1,2] \cup [3,4]$ is not clopen in $\mathbb{R}$, but $[1,2]$ and $[3,4]$ are clopen in $X$. And their union, equal to $X$ is also clopen in $X$, as any space is clopen in itself. In general in a space $X$, a finite union of clopen sets of $X$ is clopen in $X$: all unions of open sets are open, and finite unions of closed sets are closed.

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