Exercise 13.7

Proof. We claim we have the following Hasse diagram:

$T_3⊊T_1$. Inclusion is true since $U\in T_3⟹U^c$ finite, and so if we let $U^c={\{x_i\}}_{i=1}^n$ with $x_i$ in increasing order, $U={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^n(x_i,x_{i+1})$, where $x_0=-\infty ,x_{n+1}=\infty$. Inequality follows since for $(a,b)$ such that $-\infty <a,b<\infty$, $ℝ\setminus (a,b)$ is not finite.

$T_5⊊T_1$. Inclusion is clear since $(-\infty ,a)$ is of the form $(b,c)$. Inequality follows since for $(b,c)\in T_1$ and $x\in (b,c)$, there is no basis element $(-\infty ,a)\in T_5$ such that $x\in (-\infty ,a),(-\infty ,a)\subseteq (b,c)$ (if $b>-\infty$).

$T_3$ and $T_5$ are not comparable. $T_3⊈T_5$ since $ℝ\setminus \{0\}\in T_3$, but if we take $x>0$, which is in this set, there is no basis element $(-\infty ,a)\in T_5$ that contains $x$ but is contained in $ℝ\setminus \{0\}$. $T_5⊈T_3$ since ${(-\infty ,0)}^c$ is not finite.

$T_1⊊T_2$ by Lemma $13.4$.

$T_2⊊T_4$. For $(a,b)\in T_2$ and $x\in (a,b)$, $(a,x]\in T_4$ and $(a,x]\subseteq (a,b)$. For $(a,b)\setminus K\in T_2$ and $x\in (a,b)\setminus K$, we note that $x\in (1∕(n+1),c]$ where $x<c<1∕n$, $x\in (a,0]$, or $x\in (1,d]$, where $x<d<b$; in all three cases, these sets are subsets of $(a,b)\setminus K$ and are members of $T_4$. Inequality follows since for $(a,b]\in T_4$, there is no basis element $U\in T_2$ such that $b\in U,U\subset (a,b]$. □

Proof. Consider any $U\in T_3$ so that $ℝ-U$ is finite or $U=ℝ$. Clearly in the latter case $U\in T_1$ since it is a topology. In the former case $ℝ-U$ is a finite set of real numbers so that its elements can be enumerated as $\left\{x_1,x_2,\dots ,x_n\right\}$ for some $n\in {\mathbb{Z}}_{\text{+}}$ where $x_1<x_2<\cdots <x_n$. Then clearly we have that

Each of these sets is an interval $(a,b)$ or the union of such intervals. For example, the set $(-\infty ,x_1)$ can be covered by the countable union of intervals

and similarly for the interval $(x_n,\infty )$. Hence the union $U$ is an element of $T_1$ by Lemma 13.1. Since $U$ was arbitrary, this shows that $T_3\subset T_1$.

Now, clearly the interval $(-1,1)$ is in $T_1$ since it is a basis element. However, we also have that $ℝ-(-1,1)=(-\infty ,-1]\cup [1,\infty )$ is neither finite nor all of $ℝ$. Hence $(-1,1)\notin T_3$. This shows that $T_1$ cannot be a subset of $T_3$ so that $T_3⊊T_1$ as desired. □

Proof. Consider any $x\in ℝ$ and any basis element $B_5\in B_5$ containing $x$. Then $B_5=(-\infty ,a)$ where $x<a$. Let $B_1=(x-1,a)$, which is a basis element in $B_1$. Also clearly $B_1$ contains $x$ and is a subset of $B_5$. This proves that $T_5\subset T_1$ by Lemma 13.3.

Now consider $x=-1$ and basis element $B_1=(-2,0)$ in $B_1$, noting that obviously $x\in B_1$, and hence $-2<x<0$. Let $B_5$ be any element of $B_5$ containing $x$ so that $B_5=(-\infty ,a)$ where $x<a$. Clearly then $-3<-2<x<a$ so that $-3\in B_5$. However, since $-3\notin (-2,0)=B_1$, this shows that $B_5\not\subset B_1$. This suffices to show that $T_1\not\subset T_5$ by the negation of Lemma 13.3. Therefore $T_5⊊T_1$ as desired. □

Proof. First consider the set $U=\mathit{ℝnz}$ so that $U\in T_3$ since $ℝ-U=\left\{0\right\}$ is obviously finite. Now suppose that $U\in T_5$ as well. Then, since clearly $1\in U$, there must be a basis element $B_5\in B_5$ where $1\in B_5$ and $B_5\subset U$ by the definition of a topological basis. Then $B_5=(-\infty ,a)$ where $1<a$. However, since $0<1<a$ as well, it must be that $0\in B_5$, and hence $0\in U$ since $B_5\subset U$. As this clearly contradicts the definition of $U$, it has to be that $U$ is not in fact in $T_5$ so that $T_3\not\subset T_5$.

Now consider the set $U=(-\infty ,0)$, which is clearly in $T_5$ since it is a basis element. However, since $ℝ-U=[0,\infty )$ is clearly neither all of $ℝ$ nor finite, it follows that $U\notin T_3$. This shows that $T_5\not\subset T_3$, which completes the proof that the two are incomparable. □

Proof. First, consider any basis element $B_2\in B_2$ and any $x\in B_2$. Either $B_2$ is $(a,b)$ or $(a,b)-K$ for $a<b$ so that $a<x<b$ with $x\notin K$. In the former case clearly the set $B_4=(a,x]$ is in $B_4$, $x\in B_4$, and $B_4\subset B_2$. In the latter case we have the following:

Case: $x\le 0$. Then here again $B_4=(a,x]$ is in $B_4$, $x\in B_4$, and $B_4\subset B_2$ since $y\notin K$ for any $y\in B_4$ since then $a<y\le x\le 0$.

Case: $x>0$. Then let $n$ be the smallest positive integer where $n>1∕x$, which exists since ${\mathbb{Z}}_{\text{+}}$ has no upper bound and is well-ordered. It then follows that $0<1∕n<x$ and there are no integers $m$ such that $1∕n<1∕m\le x$. So let $a^{\prime }=\max <!--\; FUNCTION\; APPLICATION\; -->(a,1∕n)$ and set $B_4=(a^{\prime },x]$ so that, for any $y\in B_4$, both $a\le a^{\prime }<y\le x<b$ and $1∕n\le a^{\prime }<y<x$, and hence $y\in (a,b)$ and $y\notin K$. Therefore $y\in (a,b)-K=B_2$. Since $y$ was arbitrary, this shows that $B_4\subset B_2$, noting that also clearly $x\in B_4$ and $B_4\in B_4$.

Hence in any case it follows that $T_2\subset T_4$ from Lemma 3.13.

Now let $x=-1$ and $B_4=(-2,-1]$ so that clearly $x\in B_4$ and $B_4\in B_4$. Then let $B_2$ be any basis element in $B_2$ that contains $x$. Then we have that $B_2$ is either $(a,b)$ or $(a,b)-K$ where $a<x<b$ and $x\notin K$.

Case: $0<b$. Then $a<x=-1<0\le b$ so that $0$ is in both $(a,b)$ and $(a,b)-K$ since clearly $0\notin K$, and thus $0\in B_2$. However, clearly $0\notin (-2,-1]=B_4$.

Case: $0\ge b$. Then $a<x<(x+b)∕2<b\le 0$ so that $(x+b)∕2\in B_2$ since $(x+b)∕2$ is not in $K$. Clearly also though $(x+b)∕2\notin (-2,x]=B_4$ since $x<(x+b)∕2$.

Thus in either case we have that $B_2\not\subset B_4$. This shows the negation of Lemma 13.3 so that $T_4\not\subset T_2$. Hence $T_2⊊T_4$ as desired. □