Exercise 13.8

Proof. Let $T$ be the standard topology on $ℝ$. First, clearly $B$ is a collection of open sets of $T$ since each element is a basis element in the standard basis (i.e. an open interval). Now consider any $U\in T$ and any $x\in U$. Then there is a standard basis element $B^{\prime }=(a^{\prime },b^{\prime })$ such that $x\in B^{\prime }$ and $B^{\prime }\subset U$ since $T$ is generated by the standard basis. Then $a^{\prime }<x<b^{\prime }$ so that, since the rationals are order-dense in the reals (shown in Exercise 4.9 part (d)), there are rational $a$ and $b$ such that $a^{\prime }<a<x<b<b^{\prime }$. Let $B=(a,b)$ so that clearly $x\in B$, $B\subset B^{\prime }\subset U$, and $B\in B$. This shows that $B$ is a basis for $T$ by Lemma 13.2 since $U$ and $x\in U$ were arbitrary. □

Proof. First we must show that $C$ is a basis at all. Clearly, for any $x\in ℝ$ we have that there is an element in $C$ containing $x$, for example $[x,x+1)$. Now suppose that $C_1=[a_1,b_1)$ and $C_2=[a_2,b_2)$ are two elements of $C$ and that $x\in C_1\cap C_2$. Then obviously $a_1\le x<b_1$ and $a_2\le x<b_2$. Let $a=\max <!--\; FUNCTION\; APPLICATION\; -->(a_1,a_2)$ and $b=\min <!--\; FUNCTION\; APPLICATION\; -->(b_1,b_2)$ and $C=[a,b)$ so that clearly $C\in C$. Also clearly $a\le x<b$ since both $a_1\le x<b_1$ and $a_2\le x<b_2$, $a$ is $a_1$ or $a_2$, and $b$ is $b_1$ or $b_2$. Therefore $C$ contains $x$. Now consider any $y\in C$ so that $a_1\le a\le y<b\le b_1$ and $a_2\le a\le y<b\le b_2$ and hence $y\in C_1$ and $y\in C_2$. This shows that $C\subset C_1\cap C_2$ since $y$ was arbitrary. By definition this suffices to show that $C$ is a basis for a topology.

So let $T$ be the topology generated by $C$ and $T_l$ be the lower limit topology. Now consider $U=[x,x+1)$ where $x$ is any irrational number, for example $x=\pi$. Let $C$ be any basis element in $C$ containing $x$ so that $C=[a,b)$ where $a$ and $b$ are rational. It must be that $a\ne x$ since $a$ is rational but $x$ is not. Also, since $C$ contains $x$ it has to be that $a\le x$. So it has to be that $a<x$, but then $a\in C$ but $a\notin [x,x+1)=U$. This shows that $C$ is not a subset of $U$. Hence we have shown

This shows that $U\notin T$ by the definition of a generated topology. However, clearly we have that $U\in T_l$ since it is a lower limit basis element. This suffices to show that $T$ and $T_l$ are different topologies. □