Show that $h:X \longrightarrow \times_{i=1}^{n}{Y}$ defined by $h(x)=(f_1(x),f_2(x), \cdots, f_n(x))$ is continuous function

Question

Let $f_{i}:X \longrightarrow Y_{i}$ for $i\in \lbrace 1,2,3, \cdots,n\rbrace$ continuous functions.Show that $h:X \longrightarrow \times_{i=1}^{n}{Y}$ defined by $h(x)=(f_1(x),f_2(x), \cdots, f_n(x))$ is continuous function

Answer 1

We use induction over $i$, when $i=1$ is obviously that $f$ is continuous function, when $i=2$ we prove that the assertion is true.

Let $x\in X$ and consider $h(x)\in U_1 \times U_2$ an open set in $Y_1 \times Y_2$ since $U_1$ is open neighbourhood of $f_1(x)$ in $Y_{1}$ and $f_1$ is a continuous function there exist $V_1$ open neighbourhood of $x$ such that $f_1(V) \subset U_1$, analogue for $f_2(x)$ there exist $U_2$ open neighbourhood of $f_2(x)$ and $V_2$ open neighbourhood of $x$ such that $f(V_2) \subset U_2$.

Consider $U=U_1 \times U_2$ a neighbourhood of $h(x)$ such that exist $V=V_1 \cap V_2$ neighbourhood of $x$ such that $f(V)=F(V_1 \cap V_2)=f(V_1) \cap f(V_2) \subset V_1 \subset U_1 \subset U_1 \times U_2=U$ since $x$ was arbitrary $h$ is a continuous function as well.

Assume that the result is true for $i=n$ and we should prove for $i=n+1$.

Let $f_{i}:X \longrightarrow Y_{i}$continuous functions for $i\in \lbrace 1,2,3 \cdot, n \rbrace$ and $f_{n+1}:X\longrightarrow Y_{n+1}$ continuous function, by our assumption $h_1:X \longrightarrow \times_{i=1}^{n}Y_{i}$ is a continuous function and by hypothesis $h_{2}:X \longrightarrow Y_{n+1}$ is a continuous function, applying the case for $i=2$ we deduce that $h:X \longrightarrow (\times_{i=1}^{n}Y_i) \times Y_{n+1}$ is a continuous function, but this is equal to put $h:X \longrightarrow (\times_{i=1}^{n+1}Y_i)$ and therefore the result is true for $i\in \lbrace 1,2,3 \cdots, n \rbrace$