Exercise 16.3

Proof. We prove only the first of these as the rest follow from nearly identical arguments. Let $A=\left\{x\in ℝ\mid a<\left|x\right|<b\right\}$ and $B=(-b,-a)\cup (a,b)$ so that we must show that $A=B$.

So consider $x\in A$ so that $a<\left|x\right|<b$. If $x\ge 0$ then $\left|x\right|=x$ so that $a<x<b$ and hence $x\in (a,b)$. If $x<0$ then $\left|x\right|=-x$ so that $a<-x<b$, and thus $-a>x>-b$ so that $x\in (-b,-a)$. Thus in either case $x\in B$ so that $A\subset B$.

Now let $x\in B$ so that either $x\in (-b,-a)$ or $x\in (a,b)$. In the former case we have that $x<-a\le 0$ since $a\ge 0$ so that $\left|x\right|=-x$ and therefore

In the latter case we have that $x>a\ge 0$ so that $\left|x\right|=x$ and therefore

This shows that $B\subset A$ since $x$ was arbitrary, and thus $A=B$ as desired. □

Proof. So suppose that $U\subset Y$ is open in $X$. Then we have that $Y\cap U=U$ is also open in $Y$ by the definition of the subspace topology. □

Proof. We have from Lemma 1 that $A=(-1,-\frac{1}{2})\cup (\frac{1}{2},1)$ which is clearly the union of basis elements so that $A$ is open in $ℝ$. We also have that $A\subset Y$ so that $A$ is open in $Y$ by Lemma 2 since it is open in $ℝ$. □

Proof. By Lemmma 1 we have that $B=[-1,-\frac{1}{2})\cup (\frac{1}{2},1]$. First, consider the sets $(-2,-\frac{1}{2})$ and $(\frac{1}{2},2)$, which are clearly both basis elements and therefore open in $ℝ$. We then have that $(-2,-\frac{1}{2})\cap Y=[-1,-\frac{1}{2})$ and $(\frac{1}{2},2)\cap Y=(\frac{1}{2},1]$ so that these sets are open in $Y$ by the definition of the subspace topology. Clearly then their union $B=[-1,-\frac{1}{2})\cup (\frac{1}{2},1]$ is then also open in $Y$.

It is also easy to see that $B$ is not open in $ℝ$. For example, $-1\in B$ but for any basis element $B^{\prime }=(a,b)$ containing $-1$ we have that $a<-1<b$ so that $a<(a-1)∕2<-1<b$ and hence $(a-1)∕2\in B^{\prime }$. Clearly though $(a-1)∕2\notin B$ so that $B^{\prime }$ cannot be a subset of $B$. Thus suffices to show that $B$ is not open by the definition of the topology of $ℝ$ generated by its basis. □

Proof. By Lemmma 1 we have that $C=(-1,-\frac{1}{2}]\cup [\frac{1}{2},1)$. If $B$ is the standard basis on $ℝ$, then, by Lemma 16.1, the set $B_Y=\left\{B\cap Y\mid B\in B\right\}$ is a basis for the subspace $Y$. So consider the point $x=\frac{1}{2}$ and any basis element $B_Y\in B_Y$ containing $x$. Then we have that $B_Y=Y\cap B_X$ for some basis element $B_X=(a,b)$ in $B$, and thus $a<x<b$ since $x\in B_X$. Let $a^{\prime }=\max <!--\; FUNCTION\; APPLICATION\; -->(a,-\frac{1}{2})$ and set $y=(a^{\prime }+x)∕2$ so that

and hence $y\in (a,b)=B_X$. Also, we have

so that $y\in [-1,1]=Y$. Therefore $y\in Y\cap B_X=B_Y$. However, since $-\frac{1}{2}<y<\frac{1}{2}$, clearly $y\notin C$ so that $B_Y$ cannot be a subset of $C$. Since the basis element $B_Y\in B_Y$ was arbitrary, this suffices to show that $C$ cannot be open in $Y$ since $B_Y$ is a basis. Since also clearly $C\subset Y$, it follows from the contrapositive of Lemma 2 that $C$ is not open in $ℝ$ either. □

Proof. This follows from basically the same argument as the previous proof, again using the point $x=\frac{1}{2}$ to show that any basis element of $Y$ that contains $x$ cannot be a subset of $D$. □

Proof. First, it is trivial to show that

where we have used Lemma 1 . Now consider any $x\in E$ so that $x\in (-1,0)\cup (0,1)$ and $x\notin K$. If $x\in (-1,0)$ then clearly the basis element $(-1,0)$ contains $x$ and is a subset of $E$ since $(-1,0)\cap K=\varnothing$.

On the other hand, if $x\in (0,1)$ then $x\notin K$ so that $1∕x\notin {\mathbb{Z}}_{\text{+}}$. From this, it follows from Exercise 4.9 part (b) that there is exactly one positive integer $n$ such that $n<1∕x<n+1$. We then have that $1∕(n+1)<x<1∕n$. So let $B=(1∕(n+1),1∕n)$ so that clearly $x\in B$, $B\cap K=\varnothing$, and $B$ is a basis element of the standard topology on $ℝ$. Since $B\cap K=\varnothing$ and clearly $0<1∕(n+1)<1∕n\le 1$, it also follows that $B\subset E$.

Hence in either case there is a basis element of $ℝ$ that contains $x$ and is a subset of $E$. This suffices to show that $E$ is open in $ℝ$. Since clearly $E\subset Y$, we also clearly have that $E$ is open in $Y$ by Lemma 2 . □