I am dealing with just real line to make things little easier for me.

Suppose we have a set $X=[0,x],X'=[x,\infty)$. For the sake of argument, assume both are closed and nonempty.

Claim: By the fact that X and X' are closed and nonempty and $\mathbb{R^+}$ is connected imply $X\cap X'\not=\emptyset.$

My questions are: What are the details used here to make that claim valid? I don't quite understand probably because I don't have strong understanding of what it means for a real line to be "connected" thus how this is used here. This does make an intuitive sense if I put some numbers in, say $x=2$, then obviously the intersection of $X$ and $X'$ is the point 2. But can anybody briefly elaborate the details put into the proof of this claim?

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    $\begingroup$ Isn't it obvious the intersection equals $\{x\}?$ $\endgroup$ – zhw. Jun 20 '16 at 15:31
  • $\begingroup$ @zhw, yes it is obvious but I need to look at prove it in a rigorous way... $\endgroup$ – Frank Swanton Jun 21 '16 at 13:12
  • $\begingroup$ Here's the proof: $x \in X$ and $x\in X'$. By definition of intersection, $x\in X\cap X'.$ Therefore $X\cap X'\ne \emptyset.$ $\endgroup$ – zhw. Jun 21 '16 at 16:07
  • $\begingroup$ @zhw, haha yes. My question is actually poor worded. I will repost with a fleshed-out version, but thanks for pointing out the obvious with a poorly worded question. lol $\endgroup$ – Frank Swanton Jun 21 '16 at 21:18
  • $\begingroup$ @zhw. Of course it is obvious. However, I think that the OP posted this question as a means of understanding the concept of connectedness through this example and not solely for the sake of this example. $\endgroup$ – user119394 Jun 22 '16 at 14:35

The important point being missed here is that the $X$ and $X'$ cover $\mathbb{R^+}$ that is $X \cup X' = \mathbb{R^+}$.

A topological space is said to be connected if it cannot be partitioned into two nonempty closed sets. Therefore, if $X,X'$ were disjoint, then it would result in the partition of $\mathbb{R^+}$ into closed sets which is not possible since $\mathbb{R^+}$ is connected.

  • $\begingroup$ hey thank you for the answer. Can I ask few other questions. When you say "X and X' cover $\mathbb{R_+}$, do you mean $\mathbb{R_+}\subset(X\cup X')$? Is this the set theory definition of "cover"? Also, if we show that the union of two nonempty closed sets include the real line, is this sufficient to show $\mathbb{R_+}$ is connected or do we need to show separately the real line is connected? $\endgroup$ – Frank Swanton Jun 21 '16 at 13:15
  • $\begingroup$ @FrankSwanton Yes, you are correct regarding the definition of cover. Also, it is not enough to show that the union of two nonempty closed sets include the real line to show that $\mathbb{R^+}$ is connected. On the contrary, a space is disconnected if it can be written as the union of two disjoint closed sets. So a space $X$ is connected if and only if for all nonempty, closed $Y,Z$ that cover $X$, we have that $Y,Z$ are not disjoint. $\endgroup$ – user119394 Jun 22 '16 at 14:30

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