Let $\displaystyle f_1: X \rightarrow Y_1$ and $\displaystyle f_2: X \rightarrow Y_2$ be continuous functions. Show that $\displaystyle h: X \rightarrow Y_1\times Y_2$ defined by $\displaystyle h(x)=(f_1(x),f_2(x))$, is continuous as well.
Let $\displaystyle f_1: X \rightarrow Y_1$ and $\displaystyle f_2: X \rightarrow Y_2$ be continuous functions. Show that $\displaystyle h: X \rightarrow Y_1\times Y_2$ defined by $\displaystyle h(x)=(f_1(x),f_2(x))$, is continuous as well.
Let W be a neighborhood of $\displaystyle h(x), x \in X$ such that $\displaystyle W = U \times V $ where U is a neighborhood of $\displaystyle f_1(x)$ and V is a neighborhood of $\displaystyle f_2(x)$.
Let p be a point in X that belongs to $\displaystyle h^{-1}(U \times V)$. Then, $\displaystyle h(p) \in U \times V$ iff $\displaystyle f_1(p) \in U$ and $\displaystyle f_2(p) \in V$. Thus, $\displaystyle h^{-1}(W) = h^{-1}(U \times V) = f_1^{-1}(U) \cap f_2^{-1}(V)$. Since $\displaystyle f_1$ and $\displaystyle f_2$ are continuous and an intersection of open sets is open, $\displaystyle h^{-1}(W)$ is open. Thus, h is continuous.
Let ε>o and aεX.
Since $\displaystyle \lim_{x\rightarrow a}{f_{1}(x)}=f_{1}(a)$ and
$\displaystyle \lim_{x\rightarrow a}{f_{2}(x)} =f_{2}(a)$,then there exist:
$\displaystyle \delta_{1}>0$ and such that:
if $\displaystyle |x-a|<\delta_{1}$ and xεX ,then $\displaystyle |f_{1}(x)-f_{1}(a)|$<ε/2 for all,x............................................. .......................................1
$\displaystyle \delta_{2}>0$ and such that:
if $\displaystyle |x-a|<\delta_{2}$ and xεX, then $\displaystyle |f_{2}(x)-f_{2}(a)|$<ε/2 for all ,x................................................ ...........2.
Choose $\displaystyle \delta$ = min{$\displaystyle \delta_{1},\delta_{2}$}
Let |x-a|<δ and xεX.
then $\displaystyle |x-a|<\delta_{1}$ and $\displaystyle |x-a|<\delta_{2}$ and by (1) and (2) we have:
$\displaystyle |f_{1}(x)-f_{1}(a)|+|f_{2}(x)-f_{2}(a)|<\epsilon$
BUT.
Norm (h(x)-h(a)) = ||h(x)-h(a)|| =$\displaystyle \sqrt{(f_{1}(x)-f_{1}(a))^2 + (f_{2}(x)-f_{2}(a))^2}\leq|f_{1}(x)-f_{1}(a)| + |f_{2}(x)-f_{2}(a)|<\epsilon$
Thus $\displaystyle \lim_{x\rightarrow a}h(x) = h(a)$,for all ,a in X AND hence the function ,h is continuous over X