# Thread: Continuity with product of the range

1. ## Continuity with product of the range

Let $f_1: X \rightarrow Y_1$ and $f_2: X \rightarrow Y_2$ be continuous functions. Show that $h: X \rightarrow Y_1\times Y_2$ defined by $h(x)=(f_1(x),f_2(x))$, is continuous as well.

2. Originally Posted by Andreamet
Let $f_1: X \rightarrow Y_1$ and $f_2: X \rightarrow Y_2$ be continuous functions. Show that $h: X \rightarrow Y_1\times Y_2$ defined by $h(x)=(f_1(x),f_2(x))$, is continuous as well.
Let W be a neighborhood of $h(x), x \in X$ such that $W = U \times V$ where U is a neighborhood of $f_1(x)$ and V is a neighborhood of $f_2(x)$.
Let p be a point in X that belongs to $h^{-1}(U \times V)$. Then, $h(p) \in U \times V$ iff $f_1(p) \in U$ and $f_2(p) \in V$. Thus, $h^{-1}(W) = h^{-1}(U \times V) = f_1^{-1}(U) \cap f_2^{-1}(V)$. Since $f_1$ and $f_2$ are continuous and an intersection of open sets is open, $h^{-1}(W)$ is open. Thus, h is continuous.

3. Originally Posted by Andreamet
Let $f_1: X \rightarrow Y_1$ and $f_2: X \rightarrow Y_2$ be continuous functions. Show that $h: X \rightarrow Y_1\times Y_2$ defined by $h(x)=(f_1(x),f_2(x))$, is continuous as well.
Let ε>o and aεX.

Since $\lim_{x\rightarrow a}{f_{1}(x)}=f_{1}(a)$ and

$\lim_{x\rightarrow a}{f_{2}(x)} =f_{2}(a)$,then there exist:

$\delta_{1}>0$ and such that:

if $|x-a|<\delta_{1}$ and xεX ,then $|f_{1}(x)-f_{1}(a)|$<ε/2 for all,x............................................. .......................................1

$\delta_{2}>0$ and such that:

if $|x-a|<\delta_{2}$ and xεX, then $|f_{2}(x)-f_{2}(a)|$<ε/2 for all ,x................................................ ...........2.

Choose $\delta$ = min{ $\delta_{1},\delta_{2}$}

Let |x-a|<δ and xεX.

then $|x-a|<\delta_{1}$ and $|x-a|<\delta_{2}$ and by (1) and (2) we have:

$|f_{1}(x)-f_{1}(a)|+|f_{2}(x)-f_{2}(a)|<\epsilon$

BUT.

Norm (h(x)-h(a)) = ||h(x)-h(a)|| = $\sqrt{(f_{1}(x)-f_{1}(a))^2 + (f_{2}(x)-f_{2}(a))^2}\leq|f_{1}(x)-f_{1}(a)| + |f_{2}(x)-f_{2}(a)|<\epsilon$

Thus $\lim_{x\rightarrow a}h(x) = h(a)$,for all ,a in X AND hence the function ,h is continuous over X