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$ I \subset \Bbb R\; \text{is an interval} \Rightarrow I \subset \Bbb R\; \text{is connected}$

I am reviewing the lecture note about proving above statement.

Proof is such as below:

Suppose $I$ is a disconnected invterval.

Then $I = u \cup v$, where $u$ and $v$ are $non-empty$ $open\; subsets$ of $I$.

Thus we can find $a \in u $ and $b \in v$ with $a < b $.

Define $ c:= sup\{x\in \Bbb R\mid [a,x) \subset u \}$

Then $ a\le c \le b, hence \; c \in I$ because $a, b \in I $ and $\;\text{I is an interval}$

Clearly, $c \in \overline u\;\;$since $c$ is the limit of increasing sequence in u that converges in $I$. (*)

But $u$ is closed in $I$, since $u = v^c$, Thus $ c \in u$, and so $c<b$ and moreover since $u$ is also open in $I$, there $\exists \delta > 0$ and $B(c,\delta)\cap I \subset u $ and $c+\delta < b$.

But $[c,c+\delta) \subset [c,b] \subset I$ since $c, b \in I$ and $ I$ is an interval. (**)

Hence $[c, c+\delta) \subset B(c,\delta) \cap I \subset u$ and so $[a, c+\delta) \subset u$ which is contradict to the definition of $c$

My question is regarding (*) and (**).

  1. For the (*), how could I know that c is the limit of an increasing sequence in u and is that always contained in closure of u?

  2. For the (**), how could the fact that $c, b \in I$ and $ I$ is an interval concludes the fact that $[c,c+\delta) \subset [c,b] \subset I$ ?

Beverlie
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1 Answers1

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First quesion: Let $X = \{x\in \mathbb{R} \, \vert \, [a,x)\subseteq u\}$, so that $c = \sup(u)$. $X$ contains numbers that are arbitrarily close to $c$; given $n\in \mathbb{N}$, let $x_n\in X \cap(c-1/n, c+1/n)$. Then let $y_n = (a+nx_n)/(1+n)$. Then $y_n\in u$, and the sequence $(y_n)$ is increasing and converges to $u$.

If $S\subseteq \mathbb{R}$, and if $x$ is the limit of a sequence $(x_n)$ in $S$, then $x\in \bar{S}$. To see this, let $B(\varepsilon, x)$ be an open ball. Our goal, then, is to show that this ball intersects with $S$. Indeed, since $(x_n)\to x$, there exists $n$ such that $\vert x_n - x\vert < \varepsilon$, and so $x_n\in B(\varepsilon, x)\cap S$. This works for any $\varepsilon$, and so $x\in \bar{S}$.

Second question: Because $c+\delta < b$, it should be clear that $[c,c+\delta) \subseteq [c,b]$. Further, suppose $x\in [c,b]$. Then $c\leq x \leq b$. Since $c,b\in I$, by the definition of an interval, we have $x\in I$. This works for any $x\in [c,b]$, and so $[c,b]\subseteq I$.

florence
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