Prove $\mathbb{R}$ is connected

Question

Prove that $\mathbb{R}$ is connected. PLease i have found other ways to prove it but i want to make this way work.

Proof:

1) Strategy : If i show that a arbitrary interval is connected then i can take the colection of intervals around zero that make up $\mathbb{R}$ and have a common point So that the union is connected.

TO show that $(a,b)$ is connected ill use the fact that if it is not connected then There is an clopen in $(a,b)$ that makes up a separation with its complement in $(a,b)$ and there is not limit point of one another in any of the 2.Then ill use Least upper bound of property of that clopen set and come to a contradiction.

2) Let $(a,b)$ arbitrary open interval in $\mathbb{R}$. Suppose $\mathbb{R}$ is not connected. And also $(a,b)$ is not connected.Since it is not connected there is a clopen set $V \subseteq (a,b) $ such that it makes a separation $V \cup V^{c} =(a,b) $ . NOw since $V$ is open in $(a,b)$ $$V=K \cap (a,b)$$ and $$U=C \cap (a,b) $$ where $U=V^{c}$ and $K,C$ open in $\mathbb{R}$.

Now i have $V \cap U= \emptyset $ $$(K \cap (a,b)) \cap (C \cap (a,b)=\emptyset $$ so $$(K \cap C) \cap (a,b) =\emptyset $$

Now for the last to be true and also that my $U$ and $V$ are non empty and by drawing and trying to figure it out the $ K \cap C$ must be empty otherwise $V$ or $U$ will be empty. I can see that with drawing intervals but i cant prove it .

Hence i need to prove that $K \cap C = \emptyset $ .After i know that the specific intersection is empty things are much clear and i can take the supremum of $U$ and it will be a limit point of $U$ and be in $U$ but it will be also limit point of $V$ since every open area of it will have a non empty intersection. So it cant be a separation of $(a,b)$ and that means the only clopen sets of $(a,b)$ are itself and the empty hence it is connected and complete the proof.

Is what im trying to prove even a necessary setp? Or i can just go on with the supremum argument straight away.But i really wanna prove that intersection is empty i draw all the possibilities and it has to be empty.It is so frustrating something so easily seen not being able to write it down???

I dont want proofs using other arguments. I could easily show $R$ is path connected or myabe some other proofs. Im just stuck trying to do this one so you kinda feel my pain.

Answer 1

Your vague proof idea using suprema (Least upper bounds) is the basis for the proof of connectedness of ordered spaces:

Note that $\mathbb{R}$ is an ordered space in the sense that the topology is generated by all open sets of the form $L_a = \{x \in X: x < a\}$ and $R_a = \{x \in X: x > a\}$ , where $a \in X$. (Note that this means that the topology also contains all intervals $(a,b) = L_b \cap R_a$ as well, and the intervals with the $L_a$ and $R_a$ form a base for $X$).

An ordered topological space is connected iff it has no gaps and no jumps, which can also be stated as that it is densely ordered (for every $x < y$ in $X$ we have $z \in X$ with $x < z < y$; that's the no jumps part) and order complete (every set $A$ that is bounded above in $X$ has a least upper bound $\sup A \in X$; this is the no gaps property).

It is well-known that $\mathbb{R}$ has both of these properties, while $\mathbb{Q}$ fails the second and $\mathbb{Z}$ fails the first (and so both are disconnected ordered topological spaces).

Necessity:

Suppose $X$ has a gap $x <y$ with no points strictly in-between, then $L_y$ and $R_x$ are disjoint and non-empty, open (definition of order topology) and cover $X$ so $X$ is disconnected.

If $A$ is a set with upperbound $a_0$ but no supremum in $X$, then define $U = \{x \in X: \exists a \in A: x < a\}$, and $V = \{x \in X: \forall a \in A: a \le x\}$. $V$ is the set of upperbounds of $A$ (and this set has the property $v \in V, x > b$ then $x \in V$) and $U$ is the set of non-upperbounds of $A$ (and if $u \in U, x < u$ then $x \in U$). So by definition $U \cup V = X$.

$U$ is open, because if $u \in U$, $u < a$ for some $a \in A$ and then $u \in L_a \subseteq U$ and so $u$ is an interior point of $U$. $V$ is also open, because if $v \in B$ it's an upperbound of $A$ and there is no smallest one, so we have some smaller $b < v$ which is also an upperbound of $A$ and then $v \in R_{b} \subseteq V$, and $v$ is also an interior point of $V$. As $a_0 \in V$ and any $a \in A$ is in $U$ (or it would be a maximum, hence supemum of $A$), both sets are non-empty. So then $X$ is also disconnected.

Sufficiency

Suppose $X$ has no gaps and jumps. Assume for a contradiction that $X$ is disconnected, so $X = U \cup V$, where $U$ and $V$ are non-empty, open and disjoint. We can pick $u_0 \in U$ and $v_0 \in V$ such that $u_0 < v_0$ (we rename $U$ and $V$ if necessary). Define $U_0 = U \cap [u_0,v_0]$ and $V_0 = V \cap [u_0, v_0]$ (which are both open in $[u_0,v_0]$) and as $U_0$ is bounded above by $v_0$, $s = \sup U_0$ exists. Note that $s \in [u_0, v_0]$ ($s \le v_0$ is clear ($v_0$ is an upperbound of $U_0$, $s$ the smallest one) and $u_0 \le s$ ($s$ is an upperbound for all elements of $U_0$ so also of $u_0$). So $s \in U_0$ or $s \in V_0$.

Suppose $s \in V_0 (\subseteq V)$. Then there is an interval $(l,r)$ of $X$ such that $s \in (l,r) \subseteq V$, as $V$ is open. As $l < c$, and $l < s \le v_0$, $l$ cannot be an upperbound for $U_0$ so we have some $u \in U_0$ with $l < u$ (As $u \le s$ by definition, $u \in (l,r)$ so $u \in V$, contradiction. So $s \notin V_0$.

So then $s \in U_0$. As $U$ is open we have some interval $(l,r)$ again, such that $s \in (l,r) \subseteq U$ (clearly $r \le v_0$, or $v_0 \in U$). So $s < r$ and we find some $t$ with $s < t < r$ by the "no jumps" property. $t \in (l,r)$ so $t \in U$, but then $s$ is not even an upperbound for $U_0$ as $ t \in U_0$ and $ t > s$, contradiction. So the assumption that $X$ was disconnected was false. So a space $X$ with no gaps or jumps is connected.

Your "argument" above is much to vague ("I draw all the possibilities"..etc.; a picture (had you even included it) does not a proof make, but strict reasoning does, where pictures can assist the intuition).

Answer 2

Proving that $\mathbb{R}$ is connected is no harder than proving connectedness of any open interval.

Suppose $\mathbb{R}$ (or an open interval $(p,q)$ as well) is the disjoint union of two nonempty open sets $A$ and $B$.

Fix $a\in A$ and $b\in B$; it's not restrictive to assume $a<b$.

Consider $c=\sup\{x\in A:x<b\}$. It exists because the set $C=\{x\in A:x<b\}$ is not empty (it contains $a$) and upper bounded by $b$. Note that, by construction, $c\le b$.

There are two cases: either $c\in A$ or $c\in B$.

Suppose $c\in A$. Then $c<b$ and there exists $\delta>0$, $\delta<b-c$, such that $(c-\delta,c+\delta)\subseteq A$. In particular, $c+\delta/2\in A$ and $c+\delta/2<b$: a contradiction to $c=\sup C$.

Suppose $c\in B$. Then there exists $\delta>0$ such that $(c-\delta,c+\delta)\subseteq B$. By definition of supremum, there is $a'\in C$ with $a'>c-\delta$. But $C\subseteq A$ by assumption, so $a'\in A\cap B$, a contradiction.

Answer 3

Lemma:Every open set in $\Bbb{R}$ with respect to the Euclidean topology,can be expressed as a disjoint countable union of open intervals.

Proof:

Let $A$ be an open set in $\Bbb{R}$ and $x \in A$

We define $I_x=(c,d)$ where $$c=\inf\{a\in \Bbb{R}|(a,x) \subseteq A\}$$ $$d=\sup\{b \in \Bbb{R}|(x,b)\subseteq A\}$$

Thus $I_x$ is the largest interval that contains $x$ such that $I_x \subseteq A$

We have that $A=\bigcup_{x \in A}I_x$

Now let $x,y \in A$.

Let $I_x \cap I_y \neq \emptyset$.

Then $I_x \cup I_y \subseteq I_x$ by definition of $I_x\Rightarrow I_y \subseteq I_x$

Applying the same argument we have that $I_x \subseteq I_y\Rightarrow I_x=I_y$

So every two such intervals are disjoint.

Denote $B$ the colection of all these intervals $I$.

and let $q_I \in I$ a rational number.

Take $f:B \to \Bbb{Q}$ such that $f(I)=q_i$.This function is $1-1$ because the intervals n the collection B are disjoint.

Thus $B$ is countable.

Now assume that $\Bbb{R}$ is not connected,thus exists a nonempty clopen $A \subsetneq \Bbb{R}$

$A$ is open thus $A=\bigcup_{n=1}^{\infty}(a_n,b_n)$,a union of disjoint intervals.

Let $a_{n_0}$ be the endpoint of the interval $(a_{n_0},b_{n_0})$ in this union.

Because of the fact that $A$ is closed we have that $a_{n_0} \in A$ because $a_{n_0}$ is a limit point of $A$

Can you see the contradiction now?