Show a non-empty open and closed set in R must be equal to R [duplicate] I did this in class, and got no credit. We are now supposed to find a proof that works, can anyone help me with this? Thanks!

marked as duplicate by Joel Reyes Noche, user147263, C-Star-W-Star, colormegone, EmptyNov 3 '15 at 16:46

Here is a proof which does not assume connectedness:

Let $A$ be a non-empty, open, and closed proper subset of $\mathbb R$. Let $B = \mathbb R \setminus A$. Then $B$ is a also non-empty, open, and closed proper subset of $\mathbb R$. Let $b \in B$. Then $A$ must contain points less than $b$ or greater than $b$ (or both). (*) Suppose that $A$ contains points less than $b$. Let $$A' = A \cap (-\infty, b].$$ Note that $A'$ is closed as it is the intersection of two closed sets. Clearly $A'$ is bounded above by $b$, and so $A'$ has a least upper bound, which we will denote by $m$. Since $A'$ is closed, $m \in A'$ (and therefore $m \in A$). Furthermore, $m \neq b$ since $A$ and $B$ are complements and therefore disjoint. It follows that $m < b$. We may now conclude that the interval $(m, b]$ is a subset of $B$. This means that $m$ is a limit point of $B$, and so $m \in B$ since $B$ is closed. We now have that $m \in A$ and $m \in B$, so $A \cap B \neq \varnothing$, which is a contradiction since $B$ is the complement of $A$.

Note that if at (*) we had assumed that $A$ contains points greater than $b$, we could let $A' = [b, \infty)$ and repeat the above argument using lower bounds and again arrive at a contradiction.

Hence there exists no non-empty, open, and closed proper subset of $\mathbb R$. This statement is equivalent to saying that if $A \subset \mathbb R$ is non-empty, open, and closed, then $A = \mathbb R$.

Don't know what the proof you presented that was deemed not to work, but here is one proof, which uses the connectedness. It can be shown that any linear continuum is connected in the order topology, and $\mathbb{R}$ is clearly a linear continuum.

Now suppose that $A$ is non-empty, open and closed in $\mathbb{R}$. Then if $A\neq \mathbb{R}$, let $B=\mathbb{R}\setminus A$. Then $B$ is non-empty (since $A\neq \mathbb{R}$), open (complement of closed set), closed (complement of open set), and finally disjoint from $A$ (it's the complement of $A$). Then $A\cup B=\mathbb{R}$ is a separation of $\mathbb{R}$, giving us a contradiction.

• Path-connected $\Rightarrow$ connected uses that $\mathbb{R}$ is connected, so this is circular reasoning. – Owen Sizemore Apr 10 '14 at 1:55
• Yes, you're right. I'll fix that up. – Hayden Apr 10 '14 at 2:14

Let $a\in A$. Let $b=\sup\{x:(a, x)\subset A\}$. Now use that $A$ is closed to show if $b$ is finite then $b\in A$ then that $A$ is open to get that $(a, b+\epsilon)\subset A$ for some $\epsilon$. This shows $(a, \infty)\subset A$. Then repeat to show $(-\infty, a)\subset A$.

• Do you mean to say $(-\infty,a)$? – Cameron Williams Apr 10 '14 at 2:38
• Yes, thanks. I edited appropriately. – Owen Sizemore Apr 10 '14 at 2:41