Show that the addition function for two elements in $\mathbb{R}^\omega$ is continuous

Question

Equip $\mathbb{R}^\omega$ with the product topology. Show that the function $A\colon\mathbb{R}^\omega\times\mathbb{R}^\omega\to\mathbb{R}^\omega$ defined by

$A\bigl((a_1,a_2,\ldots,a_n,\ldots),(b_1,b_2,\ldots,b_n,\ldots)\bigr)=(a_1+b_1,a_2+b_2,\ldots,a_n+b_n,\ldots)$

is continuous.

My attempt:

$\prod U_\alpha$ is a basis element in $\mathbb{R}^\omega$ where $U_\alpha$ is open in $\mathbb{R}$ and $U_\alpha=\mathbb{R}$ except for finitely many values of $\alpha$.

We need to prove that $f^{-1}(\prod U_\alpha)$ is open in $\mathbb{R}^\omega\times\mathbb{R}^\omega$.

But this just seems too complicated.

Answer 1

The product topology is the weakest topology making every projection map continuous. This fact is key.

Answer 2

${\bf{R}}^{\omega}$ is metrizable, so we can go through by sequential characterization. For $a_{n}=(a_{1,n},a_{2,n},...)$ and $b_{n}=(b_{1,n},b_{2,n},...)$ are such that $a_{n}\rightarrow a=(a_{1},a_{2},...)$, $b_{n}\rightarrow b=(b_{1},b_{2},...)$, recourse to projection maps, one has $a_{k,n}\rightarrow a_{k}$ as $n\rightarrow\infty$, $k=1,2,...$, similar to $b_{k,n}\rightarrow b_{k}$ and hence $a_{k,n}+b_{k,n}\rightarrow a_{k}+b_{k}$, and using projections again, $A(a_{n},b_{n})=(a_{1,n}+b_{1,n},a_{2,n}+b_{2,n},...)\rightarrow(a_{1}+b_{1},a_{2}+b_{2},...)=A(a,b)$.

Answer 3

@ncmathsadist suggested a key hint. I will explain it in detail.

Wikipedia: Product topology says the universal property of the product space as follows:

Let $Y=\mathbb{R}^\omega\times\mathbb{R}^\omega$ and $X=\mathbb{R}^\omega$. Denote by $p_\alpha\colon X\to\mathbb{R}$ the canonical projection for all indices $\alpha$. Then $$ f\colon Y\to X \text{ is continuous} \iff f_\alpha=p_\alpha\circ f\colon Y\to\mathbb{R} \text{ is continuous for all $\alpha$} $$

1. By the universal property, $A\colon\mathbb{R}^\omega\times\mathbb{R}^\omega\to\mathbb{R}^\omega$ is continuous if and only if $A_\alpha=p_\alpha\circ A\colon\mathbb{R}^\omega\times\mathbb{R}^\omega\to\mathbb{R}$ is continuous for all $\alpha$.

2. Let us show that $A_\alpha\colon\mathbb{R}^\omega\times\mathbb{R}^\omega\to\mathbb{R}$ is continuous for each $\alpha$. Note that $$ p_\alpha\times p_\alpha\colon\mathbb{R}^\omega\times\mathbb{R}^\omega\to\mathbb{R}\times\mathbb{R} \quad\text{defined by}\quad (p_\alpha\times p_\alpha)(\mathbf{a},\mathbf{b})=(p_\alpha(\mathbf{a}),p_\alpha(\mathbf{b})) $$ is continuous by the universal property again. Then $A_\alpha=q\circ(p_\alpha\times p_\alpha)$ is also continuous, since the usual addition $q\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ defined by $q(a,b)=a+b$ is continuous.

Answer 4

Let. $a=(a_n)_n$ and $b=(b_n)_n.$ Let $V$ be any open subset of $\Bbb R^{\omega}$ such that $A(a,b)\in V.$ There exists $r>0$ and a finite non-empty $S\subset \omega$ such that $$V\supset \left(\prod_{n\in \omega \setminus S}\Bbb R\right)\times \left(\prod_{n\in S}I_n\right)$$ where $I_n$ is the open real interval $(-r+a_n+b_n,\;r+a_n+b_n).$

Define the open real intervals $J_n=(-r/2+a_n,\;r/2+a_n)$ and $J'_n=(-r/2+b_n,\;r/2+b_n).$

$$\text {Let }\quad U=\left(\prod_{n\in \omega \setminus S}\Bbb R\right)\times \left(\prod_{n\in S}J_n\right).$$ $$\text { Let }\quad U'=\left(\prod_{n\in \omega \setminus S}\Bbb R\right)\times \left(\prod_{n\in S}J'_n\right).$$ Then $U\times U'$ is a nbhd of $(a,b)$ in $(\Bbb R^{\omega})^2$ and $$\forall x\in U\times U'\; (\;A(x)\in V\;).$$