Vtyjkyuk

In other words, are $emptyset$ and $mathbbR$ the only open and closed sets in $mathbbR$? Why/Why not?

I tried by assuming a set is equal to its interior points and contains its limit points.

A bounded set will not do since stuff like $[1,4]$ and $5$ will not work, though that is not really proof. Help please?

Anyway, it must then be unbounded.

If $a$ is a real number then $(a,infty)$, $(-infty,a)$, $[a,infty)$ and $(-infty,a]$ don't seem to cut it so it must be $mathbbR$.

Suppose $X subset mathbbR$ is nonempty, open and closed. Let $x_0 in X$. Finally suppose that $X neq mathbbR$; then there is some $y notin X$; WLOG we can assume that $y > x_0$.

Then the set $Z = x in mathbbR : x > x_0, x notin X $ is bounded below (by $x_0$) and nonempty ($y in Z$). Therefore $inf Z = z$ exists.

• Suppose $z in X$. Then since $X$ is open, it contains an open neighborhood $(z - epsilon, z + epsilon)$. This contradicts the definition of $z = inf Z$, because there would be a sequence $z_n > z$, $|z - z_n| < frac1n$, $z_n in Z Rightarrow z_n not in X$. This is not possible, because $[z, z + epsilon) subset X$.




• Suppose $z not in X$. Then since $X$ is closed, its complement is open, therefore there is an open neighborhood $(z - epsilon, z + epsilon)$ contained in $mathbbR setminus X$. Then $z - fracepsilon2$ contradicts the $inf$ definition of $z$.

It follows that $X = mathbbR$.

You're trying to prove that $mathbbR$ is connected. You can do as follow : if a set $mathbfA$ is both open and closed, then you can check that $mathbb1_mathbfA$ (the characteristic function of $mathbfA$) is continuous, because $mathbb1_mathbfA^-1(O)$ where O is open is either $A$, its complement, the empty set or $R$ depending on $1$ and/or $0$ being in $O$, and all these set are open.

But $mathbb1_mathbfA$ only takes values $1$ or $0$, so it's easy to see that if it's continuous, then it's constant (if it is not, by the Intermediate Value Theorem, then it should also take all values between $0$ and $1$). Hence it's either always $1$ or $0$. And so $mathbfA$ is either $mathbbR$ or the empty set.

Because $mathbb R$ is connected. Asume $U$ is a nonempty proper open and closed subset of $mathbb R$, that is $V:=mathbb Rsetminus U$ is also nonempty and open.
Let $uin U$, $vin V$. Wlog. $u<v$.
Let $a=sup([u,v]cap U)$. As $[u,v]cap U$ is a nonempty bounded closed set, $ain([u,v]cap U)$. Hence $ule a<v$ (as $vnotin U$) and $U$ contains some $epsilon$ neighbourhood of $a$. But then $mina+frac12epsilon, vin Ucap[u,v]$, contradiction to $a$ being the supremum.

Let $S$ be a nonempty subset of $mathbbR$ that is both open and closed. Let's say that $S$ contains the element $s$. We want to show that no element of $mathbbR - S$ can be bigger than or less than $s$, immediately implying that it is empty.

Suppose that there are elements of $mathbbR - S$ bigger than $s$. We can construct the set $X$ of all such elements, and consider its infimum. Use the fact that $S$ is open and closed to derive a contradiction. A similar argument works to show there are no elements of $mathbbR - S$ smaller than $s$.

Note: This argument crucially uses the structure of $mathbbR$ in asserting the existence of infimums and supremums.

A space $X$ is connected if the only subsets of $X$ with empty boundary are $X$ and the empty set. Alternatively, $X$ is connected if the only subsets of $X$ which are both open and closed are $X$ and the empty set. Therefore, your first question asks if there exists any subset of $mathbb R$ that is connected. Indeed, yes, there is. An interval is connected. Your second question seems to ask if $mathbb R$ is connected. Yes, it is. For a proof, see here.

Here's yet another proof, which works by constructing a border point if $A$ is clopen nonempty proper subset of $mathbbR$:

Be $Asubsetmathbb R$ both open and closed, but neither empty nor $mathbb R$. Then there exist points $ain A$ and $binmathbb Rsetminus A$.

Now construct two sequences $(a_n)$ and $(b_n)$ as follows:

$a_0=a$, $b_0=b$. For any $n$, be $c_n=(a_n+b_n)/2$. If $c_nin A$, then $a_n+1=c_n$, $b_n+1=b_n$, else $a_n+1=a_n$, $b_n+1=c_n$.

Quite obviously for all $n$, $a_nin A$ and $b_nnotin A$. Also $lim_ntoinftyleft|a_n-b_nright| = lim_ntoinfty2^-nleft|a-bright| = 0$. Therefore there exists exactly one point $x$ so that $min(a_n,b_n)le xlemax(a_n,b_n)$ for all $n$ (nested intervals).

Since $lim_ntoinftya_n=x$ and $lim_ntoinfty b_n=x$, we have in every open neighbourhood of $x$ both points in $A$ (namely $a_n$ for sufficiently large $n$) and in the complement of $A$ (namely $b_n$ for sufficiently large $n$). Thus $x$ is a border point of $A$, in contradiction that $A$ is both open and closed, and thus cannot have any border points.