🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

TODO: Add the proof here.

TODO: Add the proof here.

We show that $0\in cl\left(A\right)$ so let $B$ be a basis element in the box topology, therefore $B=\prod (a_i,b_i)$ where $a_i<0<b_i$ then the sequence $p=(\frac{b_1}{2},\frac{b_2}{2},\frac{b_3}{2},\dots )\in B\cap A$, so that $0\in cl\left(A\right)$ as needed.

Now for the sake of contradiction suppose that $0\in seq\_cl\left(A\right)$ in othere words there exists a sequence $\left(x_n\right):{\mathbb{N}}_1\to A$ which is a sequence of positive sequences such that $x_n\to \overline{0}$. Now we will construct a specific neighborhood $U$ of $\overline{0}$, it is form $\prod (a_i,b_i)$ where $b_1={\left(x_1\right)}_1$ which is the first element of the first sequence of positive sequences that converge to $\overline{0}$, with this idea it's true that $x_1\notin U$ as the first coordinate has ${\left(x_1\right)}_1\notin (a_1,b_1)$ as it is a little too large (and in order to be in the product you must be in each). Moving in this manner we make it so that $b_i={\left(x_i\right)}_i$ so that $x_i\notin U$ but at the same time $\overline{0}\in U$ so it's not true that all but finitely many of the elements of the sequence $\left(x_n\right)$ belong to $U$, specifically none of them are!

Let $X,Y$ be metrizable spaces, with metrics given by $d,d^{\prime }$ respectively, and we define the following metric: $${\displaystyle c(x\times y,w\times z)=max(d(x,w),d^{\prime }(y,z))}$$ we know that $c$ is a metric and so now we'll show that it induces the product topology of $X\times Y$. Formally we are going to prove that the topology induced by $c$ equals the product topology by comparing basis elements

Let $x\times y$ and let $B_d(x,r_1)\times B_{d^{\prime }}(y,r_2)$ be a basis element from the product space $X\times Y$ and take $r=:min(r1,r2)$ and we claim that $${\displaystyle x\times y\in B_c(x\times y,r)\subseteq B_d(x,r_1)\times B_{d^{\prime }}(y,r_2)}$$ while that seems a bit intimidating its really just unrolling the definitions, for the inclusion suppose a point $p\times q\in B_c(x\times y,r)$ which means that $c(x\times y,p\times q)\le r$ expanding out more definitions that is $${\displaystyle max(d(x,p),d^{\prime }(y,q))=c(x\times y,p\times q)\le min(r1,r2)}$$ Now be sure to note that $d(x,p)$ is smaller than or equal to the leftmost thing there are that $r_1$ is greater than or equal to rightmost thing there. Now we want to show that $p\times q\in B_d(x,r_1)\times B_{d^{\prime }}(y,r_2)$ which is to say that $p\in B_d(x,r_1)$ which means $d(x,p)<r_1$ but this is true from what we just noted, symmetrically we also get that $q\in B_{d^{\prime }}(y,r_2)$ showing the inclusion true.

Let $x\times y$ so consider a basis element of the induced topology of $c$ it is of the form $B_c(x\times y,r)$ and we claim that $${\displaystyle x\times y\in B_d(x,r)\times B_{d^{\prime }}(y,r)\subseteq B_c(x\times y,r)}$$ to see why this holds, note that if $p\times q\in B_d(x,r)\times B_{d^{\prime }}(y,r)$ then we know that $d(x,p)<r$ and that $d^{\prime }(y,q)<r$ so that $c(x\times y,p\times q)=max(d(x,p),d^{\prime }(y,q))<r$ so $p\times q\in B_c(x\times y,r)$ as needed, therefore the two topologies are equal.

Suppose that $a<b\in \mathbb{R}$ and we'll prove that $U=d^{-1}\left((a,b)\right)\subseteq X$ is open, so let $x\times y\in U$ and consider $r=d(x,y)\in (a,b)$ which is a long way of saying $a<r<b$, where we can now find some $ϵ\in {\mathbb{R}}^{\ge 0}$ such that $a<r-2ϵ$ and $r+2ϵ<b$, while it's not necessarily obvious why we've done this we claim that $B_d(x,ϵ)\times B_d(x,ϵ)$ is a neighborhood of $x\times y$ contained in $U$, to see why consider a point $w\times z\in B_d(x,ϵ)\times B_d(x,ϵ)$ so that we know that $d(x,w)<ϵ$ and that $d(y,z)<ϵ$ so that $${\displaystyle d\backslash (w,z\backslash )\le d(w,x)+d(x,y)+d(y,z)=ϵ+r+ϵ=r+2ϵ<b}$$ in order to match with our other ineqality we can use a similar idea: $${\displaystyle d(x,y)\le d(x,w)+d(w,z)+d(z,y)}$$ and then isolate for the guy we care about $${\displaystyle d(x,y)-d(x,w)-d(z,y)\le d(z,w)}$$ but the left hand side is greater than $r-2ϵ$ and so is also greater than $a$ having the net result that $a<d(w,z)$ so all in all we know that $d(w,z)\in (a,b)$ so we can conclude that $w\times z\in d^{-1}\left((a,b)\right)=U$ and so we've proven that our open set has the property $B_d(x,ϵ)\times B_d(x,ϵ)\subseteq U$ so that $d$ is continuous.

We start with a basis element from $X$ which is some $B_d(x,ϵ)$ where $x\in X$ and $ϵ\in {\mathbb{R}}^{\ge 0}$, now since $d:X^{\prime }\times X^{\prime }\to \mathbb{R}$ was assumed continuous, then we know that it is continuous in each variable separately which we noted in class, so that the map $g\left(y\right)=d(x,y)$ is a continuous one, but then we can look at the ball again, so that $${\displaystyle B_d(x,ϵ)=\{y:d(x,y)<ϵ\}=g^{-1}\left((-ϵ,ϵ)\right)}$$ showing that $B_d(x,ϵ)$ is an open set in $X^{\prime }$ so since all basis elements are open in $X^{\prime }$ we have that $X\subseteq X^{\prime }$ as needed.

Let $\left(x_n\right)\in {\left({\mathbb{R}}^{\infty }\right)}^{\prime }$ we will prove that $\left(x_n\right)$ must converge to 0. If we take any $ϵ\in (0,1)$ and consider the open neighborhood containing $x$ which is $B_{\overline{\rho }}(x,ϵ)$ then we know that there is some $y\in B_{\overline{\rho }}(x,ϵ)\cap {\mathbb{R}}^{\infty }$ where $y\ne x$, recall that since $y\in :{\mathbb{R}}^{\infty }$ it eventually becomes zero, specifically there is some $J$ such that for all $j\ge J$ we have that $y_j=0$ but then we know that: $${\displaystyle sup\{min(|x_n-y_n|,1):n\in {\mathbb{N}}_1\}<ϵ}$$ So for any $j\ge J$ we have that $|x_j-0|=\left|x_j\right|<ϵ$, since $ϵ$ was arbitrary we've shown that $x$ converges to $0$.

Now suppose that $x\in {\mathbb{R}}^{\mathbb{N}}$ is a sequence of numbers that converge to $0$, we will now show that $x\in {\left({\mathbb{R}}^{\infty }\right)}^{\prime }$ so let $ϵ\in (0,1)$ and we consider the basis element $B_{\overline{\rho }}(x,ϵ)$. But since $x\to 0$ then we know there is some $N$ such that for all $i\ge N$ we have $x_i<ϵ$ now let's construct $y\in {\mathbb{R}}^{\infty }$ wherein we have $0<|x_1-y_1|<ϵ$ (so that $y\ne x$) then $y_i=x_i$ for $1<i<N$ and then for any $i\ge N$ set $y_i=0$ (also note that for this $i$ we know that $\left|x_i\right|<ϵ$. Also the sup over $i\in [1,\dots ,N-1]$ is equal to epsilon by our construction (coming from the first term, the rest being zeros), so we have that : $${\displaystyle sup\{min(|x_n-y_n|,1):n\in {\mathbb{N}}_1\}<ϵ}$$ which is to say that $y\in B_{\overline{\rho }}(x,ϵ)$ so that it intersects ${\mathbb{R}}^{\infty }$ in a place other than $x$ so that $x\in {\left({\mathbb{R}}^{\infty }\right)}^{\prime }$ as needed.