Exercise 18.7

Proof. Consider any basis element $B=[a,b)$, which is clearly open since basis elements are always open. We then have that the complement of this set is $C=ℝ-B=(-\infty ,a)\cup [b,\infty )$. We claim that this complement is also open so that $B$ is closed by definition. To see this, define the sets $C_n=[a-n-1,a-n+1)\cup [b+n-1,b+n+1)$ for $n\in {\mathbb{Z}}_{\text{+}}$. Clearly, each $C_n$ is open since it is the union of two basis elements. It is also trivial to show that $C={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n\in {\mathbb{Z}}_{\text{+}}}{C}_n$, which is then also open since it is a union of open sets. □

Proof. First, clearly, both $\text{∅}$ and $ℝ$ are both open and closed since they are compliments of each other and are both open by the definition of a topology. Now suppose that $U$ is a nonempty subset of $ℝ$ that is both open and closed. Suppose also that $U\ne ℝ$ so that $U⊊ℝ$ and hence there is a $y\in ℝ$ where $y\notin U$. We show that the existence of such a $U$ results in a contradiction, which of course shows the desired result since it implies that $U=ℝ$ if $U\ne \varnothing$. Since $U$ is nonempty we have that there is an $x\in U$ and it must be that $x\ne y$ since $x\in U$ but $y\notin U$.

If $x<y$ then define the set $A=\left\{z>x\mid z\notin U\right\}$. Clearly, we have that $A$ is nonempty since $y\in A$, and that $x$ is a lower bound of $A$. It then follows that $A$ has a largest lower bound $a$ since this is a fundamental property of $ℝ$. It could be that $a\in U$ or $a\notin U$. In the former case, we have that any basis element $(c,d)$ containing $a$ is not a subset of $U$. To see this, we have that $c<a<d$, which means that $d$ is not a lower bound of $A$ since $a$ is the largest lower bound. Hence there is a $z\in A$ where $d>z$. We then have $c<a\le z<d$ (noting that $a\le z$ since $a$ is a lower bound of $A$) so $z\in (c,d)$ and $z\in A$ so that $z\notin U$. Hence $(c,d)$ is not a subset of $U$, which contradicts the fact that $U$ is open since the basis element $(c,d)$ was arbitrary.

In the latter case where $a\notin U$ then it has to be that $x<a$ since $x$ is a lower bound of $A$ and $a$ is the largest lower bound (and it cannot be that $a=x$ since $x\in U$ but $a\notin U$). We clearly have that $a\in ℝ-U$, which is open since $U$ is closed. Now consider any basis element $(c,d)$ containing $a$ so that $c<a<d$. Let $b=\max <!--\; FUNCTION\; APPLICATION\; -->(x,c)$ so that $b<a$ and hence there is a real $z$ where $c\le b<z<a<d$ and hence $z\in (c,d)$. Now, since $z<a$ it has to be that $z\notin A$ since otherwise, $a$ would not be a lower bound of $A$. We also have that $x\le b<z$ so that it has to be that $z\in U$ since otherwise it would be that $z\in A$. Thus $z\notin ℝ-U$, which shows that $(c,d)$ is not a subset of $ℝ-U$ since $z\in (c,d)$. Since $(c,d)$ was an arbitrary basis element, this contradicts the fact that $ℝ-U$ is open.

It was thus shown that in either case, a contradiction arises. Analogous arguments also show contradictions when $x>y$, this time using the set $A=\left\{z<x\mid z\notin U\right\}$ and its least upper bound. Hence it has to be that $U=ℝ$, which shows the desired result. □

Proof. So suppose that $f$ is continuous from the right and consider any $a\in ℝ$. Let $V$ be neighborhood of $f(a)$ in $ℝ$. Then there is a basis element $(c,d)$ of $ℝ$ that contains $f(a)$ and is a subset of $V$. Hence $c<f(a)<d$, so let $𝜖=\min <!--\; FUNCTION\; APPLICATION\; -->[f(a)-c,d-f(a)]$ so that clearly $𝜖>0$ and if $\left|y-f(a)\right|<𝜖$, then $y\in (c,d)$ so that also $y\in V$. Now, since $f$ is continuous from the right, there is a $\delta >0$ such that $\left|f(x)-f(a)\right|<𝜖$ for every $x>a$ where $\left|x-a\right|<\delta$. So let $U=[a,a+\delta )$ which is clearly a basis element of ${ℝ}_l$ and contains $a$ so that it is a neighborhood of $a$.

Now consider any $y\in f(U)$ so that there is an $x\in U$ where $y=f(x)$. If $x=a$ then clearly $\left|f(x)-f(a)\right|=\left|f(a)-f(a)\right|=\left|0\right|=0<𝜖$ so that $f(x)\in V$. If $x\ne a$ then it has to be that $x>a$ and also that $\left|x-a\right|=x-a<\delta$ since $U=[a,a+\delta )$. It then follows that $\left|f(x)-f(a)\right|<𝜖$ so that again $f(x)\in V$. Hence in both cases $y=f(x)\in V$, which shows that $f(U)\subset V$ since $y$ was arbitrary. We have thus shown part (4) of Theorem 18.1, from which the topological continuity of $f$ follows. □

Proof. First, it was shown in Theorem 18.2 part (a) that constant functions are always continuous regardless of the topologies. Hence we must show that any continuous function from $ℝ$ to ${ℝ}_l$ is constant. So suppose that $f$ is such a function. Now consider any real $x$ where $x\ne 0$. Clearly, if $f(x)=f(0)$ then $f$ is a constant function since $x$ was arbitrary. So suppose that this is not the case so that $f(x)\ne f(0)$. Without loss of generality, we can assume that $f(0)<f(x)$. So consider the basis element $B=[f(0),f(x))$ of ${ℝ}_l$, which clearly contains $f(0)$ but not $f(x)$.

Since $f$ is continuous and $B$ is both open and closed by Lemma 1 , it follows from the definition of continuity and from Theorem 18.1 part (3) that $f^{-1}(B)$ must be both open and closed in $ℝ$. However, the only sets that are both open and closed in $ℝ$ are $\text{∅}$ and $ℝ$ itself by Lemma 2 . Thus either $f^{-1}(B)=\varnothing$ or $f^{-1}(B)=ℝ$. It cannot be that $f^{-1}(B)=\varnothing$ since we have that $f(0)\in B$ so that $0\in f^{-1}(B)$. Hence it must be that $f^{-1}(B)=ℝ$, but then we would have $x\in f^{-1}(B)$ so that $f(x)\in B$, which we know it not the case. We, therefore, have a contradiction so that it must be that $f(x)=f(0)$ so that $f$ is constant. □

Proof. First, we show that such functions are in fact continuous. So suppose that $f$ is continuous and non-decreasing from the right and consider any real $x$. Let $V$ be any neighborhood of $f(x)$ so that there is a basis element $B=[c,d)$ containing $f(x)$ such that $B\subset V$. Let $𝜖=d-f(x)$ so that $𝜖>0$ since $f(x)<d$. Hence there is a $\delta >0$ such that $x<y<x+\delta$ implies that $\left|f(y)-f(x)\right|<𝜖$ and $f(y)\ge f(x)$. We then have that $U=[x,x+\delta )$ is a basis element and therefore an open set of ${ℝ}_l$ that contains $x$. Consider any $z\in f(U)$ so that $z=f(y)$ for some $y\in U$. Then $x\le y<x+\delta$ so that $z=f(y)\ge f(x)$ and $\left|z-f(x)\right|=\left|f(y)-f(x)\right|<𝜖$. It then follows that $0\le z-f(x)<𝜖$ so that $c\le f(x)\le z<f(x)+𝜖=d$, and hence $z\in [c,d)=B$. Thus also $z\in V$ since $B\subset V$. This shows that $f(U)\subset V$ since $z$ was arbitrary, and hence that $f$ is continuous by Theorem 18.1.

Now we show that a continuous function must be continuous and non-decreasing from the right by showing the contrapositive. So suppose that $f$ is not continuous and non-decreasing from the right. Then there exists a real $x$ and an $𝜖>0$ such that, for any $\delta >0$, there is a $x\le y<x+\delta$ where $f(y)<f(x)$ or $\left|f(y)-f(x)\right|\ge 𝜖$. Clearly, we have that $V=[f(x),f(x)+𝜖)$ is basis element and therefore open set of ${ℝ}_l$ that contains $f(x)$. Consider any neighborhood $U$ of $x$ so that there is a basis element $B=[a,b)$ containing $x$ where $B\subset U$. Then $x<b$ so that $\delta =b-x>0$. It then follows that there is a $x\le y<x+\delta =b$ such that $f(y)<f(x)$ or $\left|f(y)-f(x)\right|\ge 𝜖$. Clearly, we have that $y\in B$ so that also $y\in U$ and $f(y)\in f(U)$. However, if $f(y)<f(x)$ then clearly $f(y)\notin V$. On the other hand, if $f(y)\ge f(x)$ then it has to be that $\left|f(y)-f(x)\right|\ge 𝜖$. Then we have that $f(y)-f(x)\ge 0$ so that $f(y)-f(x)=\left|f(y)-f(x)\right|\ge 𝜖$, and hence $f(y)\ge f(x)+𝜖$ so that again $f(y)\notin V$. This suffices to show that $f(U)$ is not a subset of $V$, which shows that $f$ is not continuous by Theorem 18.1 since $U$ was an arbitrary neighborhood of $x$. □