Exercise 21.4

Proof. Suppose that $x$ is any element of $\{{ℝ}_l$. Define the sets $U_n=[x,x+1∕n)$ for $n\in {\mathbb{Z}}_{\text{+}}$. Clearly, this is a countable collection of neighborhoods of $x$ since each $U_n$ is a basis element of $\{{ℝ}_l$ and $x\in U_n$. Now consider any other neighborhood $U$ of $x$ so that there is a basis element $B=[a,b)$ that contains $x$ and $B\subset U$. Thus of course $a\le x<b$. There is clearly a positive integer $N$ large enough that $N>1∕(b-x)$, noting that $b-x>0$ since $x<b$. Now consider any $y\in U_N=[x,x+1∕N)$ so that $x\le y<x+1∕N$. Then we have

so that $y\le x+1∕N<b$. As we also clearly have $a\le x\le y$ it follows that $y\in [a,b)=B\subset U$. Thus $U_N\subset U$ since $y$ was arbitrary. This shows that $\{{ℝ}_l$ satisfies the first countability axiom since $U$ and $x$ were arbitrary.

Now recall that the ordered square is the set $I\times I$ where $I=[0,1]$ with the dictionary order topology. In what follows let $\prec$ denote the dictionary order on $I\times I$. So suppose that $x=x_1\times x_2\in I\times I$ so that of course $0\le x_1\le 1$ and $0\le x_2\le 1$. Now define the following

for $n\in {\mathbb{Z}}_{\text{+}}$, which are all well defined since it is never the case that $x_2<0$ or $x_2>1$. Also define $a_n=a_{n,1}\times a_{n,2}$ and $b_n=b_{n,1}\times b_{n,2}$ for $n\in {\mathbb{Z}}_{\text{+}}$. Lastly, define the sets

for $n\in {\mathbb{Z}}_{\text{+}}$, noting that the intervals are in the dictionary order so that these are basis elements of the dictionary order topology and so are open.

As it is not obvious with all the different cases going on, we now show that $x\in U_n$ for every $n\in {\mathbb{Z}}_{\text{+}}$. So consider any $n\in {\mathbb{Z}}_{\text{+}}$.

Case: $x_2=0$. Then we have $b_{n,1}=x_1$ and $x_2<\min <!--\; FUNCTION\; APPLICATION\; -->\left\{x_2+1∕n,1\right\}=b_{n,2}$ since $x_2=0<1$ and clearly $x_2<x_2+1∕n$. This shows that $x\prec b_n$.

Case: $x_1=0$. Then we clearly have that $x_1-1∕n<x_1=0$, and hence $a_{n,1}=\max <!--\; FUNCTION\; APPLICATION\; -->\left\{x_1-1∕n,0\right\}=0=x_1$. Also clearly $a_{n,1}=0=x_2$ since $x_2=0$. Hence $a_n=0\times 0=x$ so that $a_n\preccurlyeq x$ is true. This shows that $x\in [a_n,b_n)=U_n$.

Case: $x_1>0$. Then we have $x_1-1∕n<x_1$ and $0<x_1$ so that $a_{n,1}=\max <!--\; FUNCTION\; APPLICATION\; -->\left\{x_1-1∕n,0\right\}<x_1$. Therefore $a_n\prec x$ so that $x\in (a_n,b_n)=U_n$.

Case: $x_2=1$. Then we have $a_{n,1}=x_1$ and $a_{n,2}=\max <!--\; FUNCTION\; APPLICATION\; -->\left\{x_2-1∕n,0\right\}<x_2$ since $x_2=1>0$ and clearly $x_2>x_2-1∕n$. This shows that $a_n\prec x$.

Case: $x_1=1$. Then $x_1+1∕n>x_1=1$ so that $b_{n,1}=\min <!--\; FUNCTION\; APPLICATION\; -->\left\{x_1+1∕n,1\right\}=1=x_1$. Likewise we have that $x_2+1∕n>x_2=1$ so that $b_{n,2}=\min <!--\; FUNCTION\; APPLICATION\; -->\left\{x_2+1∕n,1\right\}=1$. Thus $b_n=1\times 1=x$ and hence $x\preccurlyeq b_n$ is true. Therefore $x\in (a_n,b_n]=U_n$.

Case: $x_1<1$. Then we have $x_1+1∕n>x_1$ and $1>x_1$ so that $b_{n,1}=\min <!--\; FUNCTION\; APPLICATION\; -->\left\{x_1+1∕n,1\right\}>x_1$. Therefore $x\prec b_n$ so that $x\in (a_n,b_n)=U_n$.

Case: $0<x_2<1$. Then $a_{n,1}=x_1$, and $x_2-1∕n<x_2$ and $0<x_2$ so that $a_{n,2}=\max <!--\; FUNCTION\; APPLICATION\; -->\left\{x_2-1∕n,0\right\}<x_2$. This shows that $a_n\prec x$. Similarly $b_{n,1}=x_1$, and $x_2+1∕n>x_2$ and $1>x_2$ so that $b_{n,2}=\min <!--\; FUNCTION\; APPLICATION\; -->\left\{x_2+1∕n,1\right\}>x_2$. This shows that $x\prec b_n$. Therefore we have $x\in (a_n,b_n)=U_n$.

Hence $x\in U_n$ in all of the exhaustive cases so that ${\left\{U_n\right\}}_{n\in {\mathbb{Z}}_{\text{+}}}$ is a countable collection of neighborhoods of $x$.

Lastly consider any other neighborhood of $U$ of $x$ in the dictionary order topology. Then there is a basis element $B$ of the dictionary order topology such that $x\in B\subset U$. Then we have that either $B=[c,d)$ where $c=0$, $B=(c,d)$, or $B=(c,d]$ where $d=1$ (where we denote $1=1\times 1$). Now we set $N_a,N_b\in {\mathbb{Z}}_{\text{+}}$ based on the different cases we might have. First, we know that no matter what we have $c\preccurlyeq x$ since $x\in B$. We therefore have:

Case: $c_1<x_1$. If $x_2>0$ then set $N_a=1$, and otherwise $x_2=0$ and there is an $N_a$ large enough such that $N_a>1∕(x_1-c_1)$, noting that this is defined since $x_1-c_1>0$.

Case: $c_1=x_1$. Then it has to be that $c_2\le x_2$ since $c\preccurlyeq x$. If $c_2=x_2$ then it must be that $B=[c,d)$ and $c=x=0$, so just set $N_a=1$. On the other hand, if $c_2<x_2$, then there must be an $N_a$ large enough such that $N_a>1∕(x_2-c_2)$, noting that $x_2-c_2>0$.

Now we set $N_b$ in an analogous way, noting that we have $x\preccurlyeq d$ no matter what since $x\in B$:

Case: $x_1<d_1$. If $x_2<1$ then simply set $N_b=1$, and otherwise $x_2=1$ and there is an $N_b$ large enough such that $N_b>1∕(d_1-x_1)$, noting that $d_1-x_1>0$.

Case: $x_1=d_1$. Then it must be that $x_2\le d_2$ since $x\preccurlyeq d$. If $x_2=d_2$ then it has to be that $B=(c,d]$ and $x=d=1$, so just set $N_b=1$. On the other hand, if $x_2<d_2$, then there is an $N_b$ large enough such that $N_b>1∕(d_2-x_2)$, noting that $d_2-x_2>0$.

Now let $N=\max <!--\; FUNCTION\; APPLICATION\; -->\left\{N_a,N_b\right\}$, and we claim that $U_N\subset B$ in every case. We have again know that $c\preccurlyeq x$ so that we have:

Case: $c_1<x_1$. If $x_2>0$ then $c_1<x_1=a_{N,1}$. If $x_2=0$ then we have $N\ge N_a>1∕(x_1-c_1)$, from which it readily follows that $c_1<x_1-1∕n\le a_{N,1}$. Thus either way we have $c_1<a_{N,1}$ so that $c\prec a_N$.

Case: $c_1=x_1$. If also $c_2=x_2$ then again it has to be that $B=[c,d)$ and $c=x=0$. In this case it was established above that $U_N=[a_N,b_N)$ and $a_N=0$. So we have here that $c=0\preccurlyeq 0=a_N$. On the other hand if $c_2<x_2$ then $N\ge N_a>1∕(x_2-c_2)$, from which it follows that $c_2<x_2-1∕n\le a_{N,2}$. Also $c_1=x_1=a_{N,1}$ so that $c\prec a_N$ since $0\le c_2<x_2$.

We also of course again have that $x\preccurlyeq d$ so that

Case: $x_1<d_1$. If $x_2<1$ then $b_{N,1}=x_1<d_1$. If $x_2=1$ then we have $N\ge N_b>1∕(d_1-x_1)$, from which it follows that $b_{N,1}\le x_1+1∕N<d_1$. Hence either was $b_{N,1}<d_1$ so that $b_N\prec d$.

Case: $x_1=d_1$. If also $x_2=d_2$ then again it has to be that $B=(c,d]$ and $d=x=1$. In this case it was established above that $U_N=(a_N,b_N]$ and $b_N=1$. So we have here that $b_N=1\preccurlyeq 1=d$. On the other hand if $x_2<d_2$ then we have $N\ge N_b>1∕(d_2-x_2)$, from which it readily follows that $b_{N,2}\le x_2+1∕N<d_2$. We also have $b_{N,1}=x_1=d_1$ so that $b_N\prec d$ since $x_2<d_2\le 1$.

Therefore in every case we have that $c\prec a_N$ except when $c=x=a_N=0$ so that $U_N=[0,b_N)$ and $B=[0,d)$. Similarly we always have $b_N\prec d$ except when $b_N=x=d=1$ so that $U_N=(a_N,1]$ and $B=(c,1]$. When $c=x=a_N=0$ we cannot have $x=1$, so that $b_N\prec d$. Analogously, when $b_N=x=d=1$ it cannot be that $x=0$, and hence $c\prec a_N$. Otherwise we have $U_N=(a_N,b_N)$, $c\prec a_N$, and $b_N\prec d$ so that in every case $x\in U_N\subset B\subset U$. Since $U$ was an arbitrary neighborhood and $x$ was also arbitrary, this shows that $I\times I$ satisfies the first countability axiom as desired. □