# Proof: Continuity in $\mathbb R$ implies continuity in $\mathbb R^n$?

I am trying to prove this statement about continuity on the reals:]

Let $f:\mathbb R\to\mathbb R$ be continuous using the Euclidean distance function. Then the function $g:\mathbb R^n \to\mathbb R^n$ defined by $$g(x_1,x_2,...,x_n)=(f(x_1),f(x_2),...,f(x_n))$$ is continuous in Euclidean $n$-space.

Here is my proof:

Since $f$ is continuous, for any given value of $\epsilon\gt 0$, there exists $\delta\gt 0$ such that $$|a-b|\lt \delta\implies|f(a)-f(b)|\lt \epsilon$$ I will use $\delta[\epsilon]$ to represent such a value of $\delta$.

Now suppose that $\epsilon'\gt 0$ is given. Let $$\delta'=\delta\bigg[\frac{\epsilon'}{\sqrt n}\bigg]$$ Then suppose we have that $$\sqrt{\sum_{k=1}^n (a_k-b_k)^2}\lt \delta'$$ $$\sum_{k=1}^n (a_k-b_k)^2\lt \delta'^2$$ and, for each $k$, $$(a_k-b_k)^2\lt \delta'^2$$ $$|a_k-b_k|\lt \delta'$$ $$|a_k-b_k|\lt \delta\bigg[\frac{\epsilon'}{\sqrt n}\bigg]$$ and, by the way we defined $\delta[\epsilon']$, this implies $$|f(a_k)-f(b_k)|\lt \frac{\epsilon'}{\sqrt n}$$ $$n(f(a_k)-f(b_k))^2\lt \epsilon'^2$$ $$(f(a_k)-f(b_k))^2+(f(a_k)-f(b_k))^2+...+(f(a_k)-f(b_k))^2\lt \epsilon'^2$$ where the LHS of the above statement contains $n$ copies of $(f(a_k)-f(b_k))^2$. Since this is true for all $k$, even the $k$ for which $(f(a_k)-f(b_k))^2$ is maximized, we have that $$(f(a_1)-f(b_1))^2+(f(a_2)-f(b_2))^2+...+(f(a_n)-f(b_n))^2\lt \epsilon'^2$$ $$\sum_{k=1}^n (f(a_k)-f(b_k))^2\lt \epsilon'^2$$ $$\sqrt{\sum_{k=1}^n (f(a_k)-f(b_k))^2}\lt \epsilon'$$ Which completes the proof.

Is this proof correct?

• Correct. However, the remark about $(f(a_k) - f(b_k))^2$ maximization is unnecessary, just add up the inequalities $(f(a_k)-f(b_k))^2 < \epsilon'^2/n$ for $k=1, 2, \ldots, n$. – Adayah Jul 31 '17 at 14:32
• Your $k$'s turned into $n$'s at the end, – zhw. Jul 31 '17 at 14:34
• @Adayah Okay, thanks! – Franklin Pezzuti Dyer Jul 31 '17 at 14:35
• @zhw Whoops, thank you! – Franklin Pezzuti Dyer Jul 31 '17 at 14:36