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I am reading "Principles of Mathematical Analysis 3rd Edition" by Walter Rudin.

There is the following theorem on p.42:

Theorem 2.47
A subset $E$ of the real line $\mathbb{R}^1$ is connected if and only if it has the following property: If $x \in E, y \in E$, and $x < z < y$, then $z \in E$.

Rudin didn't write the following more concrete result.

Why?

I wish if Rudin had written the following more concrete result:

A subset $E$ of the real line $\mathbb{R}^1$ has the following property: If $x \in E, y \in E$, and $x < z < y$, then $z \in E$

if and only if

(1) $E = (a, b)$ for $a, b \in \mathbb{R}$ such that $a \leq b$ or
(2) $E = [a, b]$ for $a, b \in \mathbb{R}$ such that $a \leq b$ or
(3) $E = [a, b)$ for $a, b \in \mathbb{R}$ such that $a \leq b$ or
(4) $E = (a, b]$ for $a, b \in \mathbb{R}$ such that $a \leq b$ or
(5) $E = (a, +\infty)$ for $a \in \mathbb{R}$ or
(6) $E = [a, +\infty)$ for $a \in \mathbb{R}$ or
(7) $E = (-\infty, b)$ for $b \in \mathbb{R}$ or
(8) $E = (-\infty, b]$ for $b \in \mathbb{R}$ or
(9) $E = (-\infty, +\infty)$.

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    $\begingroup$ Your theorem has nothing to do with connectedness. It complements Rudin's theorem but it is not a replacement. $\endgroup$ – Kavi Rama Murthy Feb 4 at 9:15
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Why? Because the two statements are equivalent and yours is way more work to write down. The essence of the theorem is that connected sets of reals (w.r.t. the usual topology) are precisely intervals.

Edit: I see your version of the theorem doesn't mention connectedness and is therefore false. But the above applies by correcting your statement in the obvious way.

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I would imagine it's because connectedness isn't as simple as an interval when you get into different situations, see eg. topologists curve and polish curve.

The first form of the statement gets you used to thinking outside of intervals, the second form keeps intervals fixed in your head as fundamental, it also takes up a lot of room and Rudin does tend to like everything to be neat and compact.

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