# Prove $\mathbb{R}$ is connected

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.

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).

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.

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?