Exercise 2.4.4 (Continuity preserves connectedness)

$(a)⟹(b)$ Suppose $X$ is connected. Suppose for the sake of contradiction that (b) may not hold, i.e.,

$$\neg <!--\; FUNCTION\; APPLICATION\; -->\left(\forall <!--\; FUNCTION\; APPLICATION\; -->x,y\in X:x<y⟹[x,y]\subseteq X\right)$$

$$⟺\exists <!--\; FUNCTION\; APPLICATION\; -->x,y\in X:x<y\text{ and }[x,y]⊈X$$

Thus, there exists at least one real number $z\in (x,y)$ such that $z\notin X$. Hence, consider the intervals

$$(-\infty ,z)\cap X\text{and}X\cap (z,+\infty )$$

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 $\left[(-\infty ,z)\cap X\right]\cap \left[X\cap (z,+\infty )\right]\subseteq (-\infty ,z)\cap (z,+\infty )=\varnothing$
• 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 $ℝ$. This is the case, since open intervals $V=(-\infty ,z),(z,+\infty )$ are open in $ℝ$ 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 $X$.

$(b)⟹(a)$ Suppose $X$ satisfies property (b), i.e.,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->x,y\in X:x<y⟹[x,y]\subseteq X$$

and suppose, for the sake of contradiction, that $X$ may be not connected, i.e., there exist some

(1)

non-empty,

(2)

disjoint,

(3)

open relative to $X$

sets $V$ and $W$ whose union is $X$ itself. Since $V$ and $W$ are both (1) non-empty, by double choice we pick $x\in V$ and $y\in W$. Since $V$ and $W$ are (2) disjoint, we have $x\ne y$; 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 $y$. Thus, due to the Theorem I.5.5.9. it must obtain the least lower bound

$$\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)$$

Since $\left(V\cap [x,y]\right)$ is bounded by $x$ and $y$, we in particular have

$$\begin{array}{cc}x\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)\le y\hfill & ⟹\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)\in [x,y]\hfill \\ \hfill & ⟹\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)\in X\hfill \end{array}$$

Since $X$ is the disjoint union of $V\cup W$, we are left with the two mutually exclusive cases:

1.

$\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)\in V$
On one hand, since $V$ is open relative to $X$, we can find a ball around $\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)$ which is small enough to be contained in $V$, i.e, $$\exists <!--\; FUNCTION\; APPLICATION\; -->r_1>0:B\left(\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right),r_1\right)\subseteq V$$

On the other hand, since $y\in W$ and $V\cap W=\varnothing$, we have in such case $x\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)<y$. Hence, $\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)$ is contained also in $[x,y]$; therefore,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->r_2<y-\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right):\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)+r_2\in [x,y]$$

Set $r:=\min <!--\; FUNCTION\; APPLICATION\; -->\{r_1,r_2\}$. Then the number

$$\begin{array}{cc}\hfill \sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)+r∕2\in & B\left(\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right),r\right)\subseteq V\hfill \\ \hfill \in & [x,y]\hfill \end{array}$$

Thus although $\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)+r∕2>\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)$ we have $\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)\in V\cap [x,y]$ - a contradiction to the definition of supremum.

2.

$\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)\in W$
Similarly: on one hand, since $W$ is open relative to $X$, we can find a ball around $\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)$ which is small enough to be contained in $W$, i.e, $$\exists <!--\; FUNCTION\; APPLICATION\; -->r_1>0:B\left(\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right),r_1\right)\subseteq W$$

On the other hand, since $x\in V$ and $V\cap W=\varnothing$, we have in such case $x<\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)\le y$. Hence, $\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)$ is contained also in $[x,y]$; therefore,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->r_2<\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)-x:\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)-r_2\in [x,y]$$

Set $r:=\min <!--\; FUNCTION\; APPLICATION\; -->\{r_1,r_2\}$. Then the number

$$\begin{array}{cc}\hfill \sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)-r∕2\in & B\left(\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right),r\right)\subseteq W\hfill \\ \hfill \in & [x,y],\hfill \end{array}$$

Thus, although $x<\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)-r∕2<\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)⟹\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)-r∕2\in V\cap [x,y]$ we have $\sup <!--\; FUNCTION\; APPLICATION\; -->\left(V\cap [x,y]\right)-r∕2\in W$ - 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.

$(b)⟹(c)$ Suppose that $X$ obeys the property (b), i.e.,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->x,y\in X:x<y⟹[x,y]\subseteq X$$

Suppose, for the sake of contradiction, that $X$ must not be an interval, i.e.,

$$\neg <!--\; FUNCTION\; APPLICATION\; -->\left(\exists <!--\; FUNCTION\; APPLICATION\; -->a,b\in {ℝ}^{\text{∗}}:a<b\forall <!--\; FUNCTION\; APPLICATION\; -->z\in ℝ:a<z<b⟹z\in X\right)$$

$$\forall <!--\; FUNCTION\; APPLICATION\; -->a,b\in {ℝ}^{\text{∗}}:a<b\exists <!--\; FUNCTION\; APPLICATION\; -->z\in ℝ:a<z<b\text{ and }z\notin X$$

But this is a contradiction to the property (b), since for some arbitrary $x,y\in X\subseteq ℝ\subset {ℝ}^{\text{∗}}$ the interval $[x,y]$ does not contain any $z:x<z<y$ and $z\notin [x,y]\subseteq X$.

$(c)⟹(b)$ Suppose that $X$ is an interval and assume $X=[a,b]$ (other cases are similar). Then $X=\{x\in ℝ:a\le x\le b\}$. Thus, for any $x,y\in X$ it is true that $a\le x,y\le b$. Hence, $\forall <!--\; FUNCTION\; APPLICATION\; -->z:z\in [x,y]⟹z\in [a,b]$.