# What is the general outline of the proof that the real line is connected? [closed]

I am reading many topology books and I want to understand the proofs that the real line is connected.

What is the general structure and outline showing that the real line is connected?

• This may be, perhaps, trivial, depending on the assumptions made and what can be used. For example, take $\;x,y\in\Bbb R\;$ and assume WLOG that $\;x<y\;$ . Then, by definition (or not: it depends), we get $\;[x,y]\subset\Bbb R\;$, which means (or not: it depends) $\;\Bbb R\;$ is path connected and thus connected. QED (...or not: depending on what can one assume). Commented Jul 27, 2016 at 19:17
• @DonAntonio: Using path connectivity of $\mathbb{R}$ to prove connectivity of $\mathbb{R}$ is circular reasoning, because the theorem that every path connected space is connected depends on the theorem that $\mathbb{R}$ is connected. Commented Jul 27, 2016 at 19:19
• @LeeMosher I think you may be right yet, as far as I recall, it only uses the fact that $\;[0,1]\;$ is connected, which is, perhaps, to assume less than to assume the whole line is connected. Commented Jul 27, 2016 at 19:28
• @DonAntonio Given that $\mathbb R$ and $(0, 1)$ are homeomorphic, I'm not sure how that follows. The proof that $[0, 1]$ is connected is essentially the same as that of $\mathbb R$. Commented Jul 27, 2016 at 19:39
• @AymanHourieh You think? Perhaps so...it doesn't matter, really. The above was only the sketch of an idea... Commented Jul 27, 2016 at 19:44

Suppose towards contradiction that $\mathbb{R}=U\sqcup V$, $U$ and $V$ nonempty and open.

Then there are real numbers $x\in U$ and $y\in V$ - without loss of generality let's assume $x<y$.

Now - looking at the interval $[x, y]$ - let $$z=\inf\{a\in [x, y]: a\in V\}.$$ Such a real $z$ exists, by the completeness of $\mathbb{R}$.

Now, which piece of $\mathbb{R}$ is $z$ in?

• Well, by definition $z$ is a limit point of $V$ (from the right). But since $U$ is open, $V$ is closed, so $z$ must be in $V$!

• At the same time, since $z\in V$ we must have $x<z$; so $z$ is a limit point of $U$ (from the left). So since $V$ is open, $U$ is closed, so $z$ must be in $U$!

• But $U$ and $V$ are disjoint.

• Maybe it's better to point out explicitly that $[x,z)\subseteq U$, because no element in $[x,z)$ can belong to $V$. Commented Jul 27, 2016 at 19:54
• You assume $\mathbb{R}=U\cup V$ where $U, V$ are nonempty and open, but when you try to conclude where $z$ belongs to, then said "But since $U$ is open, $V$ is closed". Why? @NoahSchweber Commented May 29, 2021 at 9:24
• @falamiw By definition, a set is closed if its complement is open. Since $U$ is open and $V$ is the complement of $U$, $V$ is closed. Commented May 29, 2021 at 13:58