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Show that $[a,b] \cup [c,d]=T$, where $a < b$, $c < d$ and $b < c$ is not connected (subset of R)

In most books (as Rudin) there are propositions that uses the fact there exists $b < x < c$ and creates $(- \infty, x)\cap T$ and $(x ,+ \infty)\cap T$. Why are those intervals open? This must be open in R or $T$?

Thanks in advance!

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  • $\begingroup$ An subset $I$ of $\Bbb R$ is connected iff it has the following property: if $a,b\in I$ and $a<c<b$, then $c\in I$. $\endgroup$ – Pedro Tamaroff Nov 30 '13 at 20:10
  • $\begingroup$ @PedroTamaroff Right, but the asker is looking for a proof of (a weaker version of) that fact. $\endgroup$ – Potato Nov 30 '13 at 20:14
  • $\begingroup$ @Potato Just saying, that is not too hard to prove, and comes in handy. =) $\endgroup$ – Pedro Tamaroff Nov 30 '13 at 20:16
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$(-\infty,x) \cap T$ and $(x,+ \infty) \cap T$ are open in the subspace topology, which is generated by taking the intersection of $T$ with all of the open subsets of $\mathbf R$. Such sets are called "open in $T$" and they need not be open in $\mathbf R$.

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Let $P = (-\infty,(b + c)/2)$ and $Q = ((b + c)/2, \infty)$. Thus $P$ and $Q$ are open subsets of $\mathbb{R}$. Let $X = [a,b]\cup[c,d]$, then $X$ is a subset of $P \cup Q$. Also, $X\cap P\cap Q = \emptyset$ since $P\cap Q = \emptyset$. And $X \cap P\neq\emptyset$, and $X\cap Q\neq\emptyset$, which is easy to verify. All of these show that the set $X$ in question is disconnected.

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  • $\begingroup$ You should not answer a question without the use of LaTex typing. $\endgroup$ – Paulo Henrique Nov 30 '13 at 21:03

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