Connectedness on the real line

Question

This is rather long proof of equivalence of the definitions of connectedness on the real line. Because I am self-learning, I would really like to know if my proof contains any flows or logical fallacies. Since it is somewhat long, I would very mucht appreciate the help. The problem is adopted from Tao (2011, p36).

Definition 2.4.1 (Connected spaces). Let $(X, d)$ be a metric space. We say that X is disconnected iff there exist disjoint non-empty open sets V and W in X such that $V \cup W = X$. We say that $X$ is connected iff it is non-empty and not disconnected.

Theorem 2.4.5. Let $(\mathbb{R},|\cdot|)$ be the real line. Let $X\subseteq \mathbb{R}$. Then the following statements are equivalent.

1. $X$ is connected.
2. Whenever $x, y \in X$ and $x < y$, the interval $[x, y]$ is also contained in $X$.
3. X is an interval

$(1) \implies (2)$ Suppose $X$ is connected. Suppose for the sake of contradiction that (b) is not true, i.e., $$\neg \left( \forall x,y \in X: x<y \implies [x,y] \subseteq X \right)$$ $$\iff \exists x,y \in X: x<y \text{ and } [x,y] \nsubseteq X $$ Thus, there exists at least one real number $z\in (x,y)$ such that $z\notin X$. Hence, consider the intervals $$(-\infty,z)\cap X \quad \text{and} \quad X \cap (z,+\infty)$$ These sets are

• non-empty, since $x < z < y$, and hence at least $x \in (-\infty,z)\cap X$ and $y \in X \cap (z,+\infty)$
• disjoint, since $\left[(-\infty,z)\cap X\right] \cap \left[X \cap (z,+\infty) \right] = (-\infty,z)\cap (z,+\infty) = \emptyset$
• open relative to $X$, since by Theorem 1.3.4. $(-\infty,z)\cap X$ is relatively open with respect to $X$ if and only if $(-\infty,z)\cap X = V \cap X$ for some set $V \subseteq X$ which is open in $\mathbb{R}$. This is the case, since open intervals $V=(-\infty,z),(z,+\infty)$ are open in $\mathbb{R}$ due to the Lemma I.9.1.12.

Thus, using the negation of (b), we have found two non-empty disjoint sets $(-\infty,z)\cap X \text{ and } X \cap (z,+\infty)$, which are open relative to $X$, and which cover $X$. But this is a contradiction to the connectedness assumption on $X$.

$(2) \implies (1)$ Suppose $X$ satisfies property (b), i.e., $$\forall x,y \in X: x<y \implies [x,y] \subseteq X$$ and suppose, for the sake of contradiction, that $X$ may be not connected, i.e., there exist some non-empty, disjoint, open relative to $X$ sets $V$ and $W$ whose union is $X$ itself. Since $V$ and $W$ are non-empty, by double choice we pick $x\in V$ and $y\in W$. Since $V$ and $W$ are disjoint, we have $x\neq y$; hence either $x>y$ or $x<y$. Without loss of generality assume $x<y$. At this point we employ property (b) and see that $$[x,y] \subseteq X$$ Now consider the set $V \cap [x,y]$. This set is non-empty, since it at least contains $x\in V,[x,y]$. Also $V \cap [x,y] \subseteq [x,y]$; hence $V \cap [x,y]$ is bounded by $x$ and $y$. Due to the Theorem I.5.5.9. it must obtain the least lower bound $$\sup \left( V \cap [x,y] \right)$$ Since $\left( V \cap [x,y] \right)$ is bounded by $x$ and $y$, we in particular have $$\begin{array}{ll} x\leq \sup \left( V \cap [x,y] \right) \leq y &\implies \sup \left( V \cap [x,y] \right) \in [x,y]\\ &\implies \sup \left( V \cap [x,y] \right) \in X \end{array}$$ Since $X$ is the disjoint union of $V \cup W$ we have two mutually exclusive cases:

1. $\sup \left( V \cap [x,y] \right) \in V$
   On one hand, since $V$ is open relative to $X$, we can find a ball around $\sup \left( V \cap [x,y] \right)$ which is small enough to be contained in $V$, i.e, $$\exists r_1>0: \quad B\left(\sup \left( V \cap [x,y] \right), r_1 \right) \subseteq V$$ On the other hand, since $y \in W$ and $V\cap W = \emptyset$, we have in such case $x\leq \sup \left( V \cap [x,y] \right) < y$. Hence, $\sup \left( V \cap [x,y] \right)$ is contained also in $[x,y]$; therefore, $$\forall r_2 < y-\sup \left( V \cap [x,y] \right): \quad \sup \left( V \cap [x,y] \right)+ r_2 \in [x,y]$$ Set $r := \min\{r_1,r_2 \}$. Then the number $$\begin{array}{rll} \sup \left( V \cap [x,y] \right)+r/2 \in &B\left(\sup \left( V \cap [x,y] \right), r \right) \subseteq V \\ \in &[x,y] \end{array}$$ Thus although $\sup \left( V \cap [x,y] \right)+r/2 > \sup \left( V \cap [x,y] \right)$ we have $\sup \left( V \cap [x,y] \right) \in V\cap [x,y]$ - a contradiction to the definition of supremum.
2. $\sup \left( V \cap [x,y] \right) \in W$
   Similarly: on one hand, since $W$ is open relative to $X$, we can find a ball around $\sup \left( V \cap [x,y] \right)$ which is small enough to be contained in $W$, i.e, $$\exists r_1>0: \quad B\left(\sup \left( V \cap [x,y] \right), r_1 \right) \subseteq W$$ On the other hand, since $x \in V$ and $V\cap W = \emptyset$, we have in such case $x< \sup \left( V \cap [x,y] \right) \leq y$. Hence, $\sup \left( V \cap [x,y] \right)$ is contained also in $[x,y]$; therefore, $$\forall r_2 < \sup \left( V \cap [x,y] \right)-x: \quad \sup \left( V \cap [x,y] \right)- r_2 \in [x,y]$$ Set $r := \min\{r_1,r_2 \}$. Then the number $$\begin{array}{rll} \sup \left( V \cap [x,y] \right)-r/2 \in &B\left(\sup \left( V \cap [x,y] \right), r \right) \subseteq W\\ \in &[x,y], \end{array}$$ Thus, although $x<\sup \left( V \cap [x,y] \right)-r/2<\sup \left( V \cap [x,y] \right) \implies \sup \left( V \cap [x,y] \right)-r/2 \in V\cap [x,y]$ we have $\sup \left( V \cap [x,y] \right)-r/2 \in W$ - a contradiction to the assumption that $V$ and $W$ are disjoint.

Thus in either case we obtain a contradiction, which means that $X$ cannot be disconnected, and must therefore be connected.

Thus in either case we obtain a contradiction, which means that $X$ cannot be disconnected, and must therefore be connected.

$(2) \implies (3)$ Suppose that $X$ obeys the property (b), i.e., $$\forall x,y \in X: x<y \implies [x,y] \subseteq X$$ Suppose, for the sake of contradiction, that $X$ is not an interval, i.e., $$\neg \left( \exists a,b \in \mathbb{R}^{*}: a<b \quad \forall z \in \mathbb{R}: \quad a < z <b \implies z \in X\right)$$ $$\forall a,b \in \mathbb{R}^{*}: a<b \quad \exists z \in \mathbb{R}: \quad a < z <b \text{ and } z \notin X$$ But this is a contradiction to the property (b), since for some arbitrary $x,y \in X \subseteq \mathbb{R} \subset \mathbb{R}^{*}$ the interval $[x,y]$ does not contain any $z: x<z<y$ and $z\notin [x,y] \subseteq X$.

$(3) \implies (2)$ Suppose that $X$ is an interval and assume $X=[a,b]$ (other cases are similar). Then $X = \{ x\in \mathbb{R}: a \leq x \leq b\}$. Thus, for any $x,y \in X$ it is true that $a\leq x,y \leq b$. Hence, $\forall z: z\in [x,y] \implies z \in [a,b]$.