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.

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- Apr 4th 2009, 08:40 AMAndreametContinuity with product of the range
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.

- Apr 4th 2009, 09:45 PMaliceinwonderland
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. - Apr 6th 2009, 04:44 AMxalk
**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