$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

$⟹$ Suppose that $x_n\to x$, and fix $\beta \in A$ and let $U$ be a neighborhood of ${\pi }_{\beta }\left(x\right)$ and consider $\prod C_{\alpha }$ where $C_{\alpha }=X_{\alpha }$ whenever $\alpha \ne \beta$ and $C_{\alpha }=U$ when $\alpha =\beta$, therefore we know that $x\in B:=\prod C_{\alpha }$ since this is a product of open sets where all but finitely many are the whole space then $B$ is a basis element of the product topology and thus is an open set containing $x$ therefore by assumption there exists an $N$ such that for all $k\ge N$ we have that $x_k\in B$ meaning that ${\pi }_{\beta }\left(x_k\right)\in U$, therefore ${\pi }_{\alpha }\left(x_k\right)\to {\pi }_{\alpha }\left(x\right)$ as needed.

Suppose that ${\pi }_{\alpha }\left(x_k\right)\to {\pi }_{\alpha }\left(x\right)$ for each $\alpha$, and let's prove that $x_n\to x$ so let $U$ be an open neighborhood of $x$. Since $x\in \prod X_{\alpha }$ then we know that there exists a basis element $B$ from the product topology such that $x\in B\subseteq U$ recall that $B=\prod V_{\alpha }$ where there exists some finite set $F\subseteq A$ such that $V_{\alpha }$ is an open set in $X_{\alpha }$ whenever $\alpha \in F$ and $V_{\alpha }=X_{\alpha }$ otherwise.

Since $x\in \prod V_{\alpha }$ then we know that ${\pi }_{\alpha }\left(x\right)\in V_{\alpha }$, note that $V_{\alpha }$ is an open neighborhood of ${\pi }_{\alpha }\left(x\right)$, now we restrict our attention to $\alpha \in F$ now by assumption we know that there exists some $N_{\alpha }$ such that whenever we have $j>N_{\alpha }$ then ${\pi }_{\alpha }\left(x_j\right)$, since $F$ was finite, then $N=\max \left(\{N_{\alpha }:\alpha \in F\}\right)$.

Now suppose that $s>N$ therefore we know that $x_s\in V_{\alpha }$ for every $\alpha \in F$ as constructed in the previous paragraph, additionally whenver $\alpha \in A\setminus F$ then $V_{\alpha }=X_{\alpha }$ therefore we definitely know that $x_s\in V_{\alpha }$ in that case. Therefore we've shown that $x_s\in B$ and since $B\subseteq U$ then $x_s\in U$ as needed, so it converges.

Also note that if we consider $\prod X_{\alpha }$ with the box topology then the statement is no longer true, because if we consider $x=(0,0,0,\dots )\in {ℝ}^{\infty }$ then ${\prod }_{n\in {\mathbb{N}}_1}(\frac{-1}{n},\frac{1}{n})$ is a neighborhood of $x$, and consider the sequence of elements from $\prod X_{\alpha }$ where we define $x_i=(\frac{1}{i},\frac{1}{i},\dots )$ for each $i\in {\mathbb{N}}_1$ then ${\pi }_j\left(x_n\right)=(1,\frac{1}{2},\frac{1}{3},\dots )$ converges to $0={\pi }_j\left(x\right)$ but then its clear that $\left(x_n\right)$ does not converge to $x$ because for any $k\in {\mathbb{N}}_1$ there is a point at which $U$ gets too small and excludes elements from $x_k$, in other words $x_k\notin U$.

We will show that $\overline{{ℝ}^{\infty }}={ℝ}^{\infty }$ by showing that the set is closed by by following this corollary. So let $\left(x_n\right)\notin {ℝ}^{\infty }$ then that means that there are infinitely many $i\in {\mathbb{N}}_1$ such that $x_i\ne 0$, for each such $x_i$ since it is an element of $R$ then we can construct an open set around $x_i$ that does not contain $0$ by using setting $\delta =\frac{x_i}{2}$ and considering the interval $(x_i-\delta ,x_i+\delta )$ let's denote that set by $U_i$, we can expand this definition for all $i\in {\mathbb{N}}_0$ by considering the places at which $x_i=0$, for those places we will set $U_i=ℝ$, then note that for any $y\in \prod U_i$ then $y$ also has infinitely many entries that are non-zero, thus $y\notin {ℝ}^{\infty }$, this proves that ${ℝ}^{\infty }\cap \prod U_i=\varnothing$ meaning that there exists a neighborhood which doesn't intersect ${ℝ}^{\infty }$ at all and thus since $\left(x_n\right)$ was an arbitrary element of ${ℝ}^{ℝ}\setminus {ℝ}^{\infty }$ then no point there is a limit point, therefore the corollary tells us that ${ℝ}^{\infty }$ is closed.

Now we do the same but for the product topology but we only focus on elements outside and now let $x\in {ℝ}^{ℝ}\setminus {ℝ}^{\infty }$ , and let $B=\prod U_n$ be a basis element containing $x$ therefore $U_n=ℝ$ for all but finitely many indices, and therefore there exists some $N\in {\mathbb{N}}_1$ such that for any $n\ge N$ we have that $U_n=ℝ$ we now construct $y$ such that $y_n=x_n$ for all $n<N$ and $y_n=0$ whenever $n\ge N$ we observe that $x\ne y$ as $x$ has infinitely many elements of its sequence that are non-zero, whereas $y$ only has finitely many that are non-zero which also tells us that $y\in {ℝ}^{\infty }$ which shows that $y\in B\cap {ℝ}^{\infty }\ne \varnothing$ therefore we know that $x\in {\left(R^{\infty }\right)}^{\prime }$ which shows that every point outside ${ℝ}^{\infty }$ is a limit point, so we conclude that $\overline{{ℝ}^{\infty }}={ℝ}^{ℝ}$, as needed.