Let $X$ and $Y$ be topological spaces; $X\times Y$ carries the product topology. Can someone provide me with an easy argument that for $f\in C(X,\mathbb{R})$ and $g\in C(X,\mathbb{R})$ the mapping $$ h\colon X\times Y \to \mathbb{R},\ (x,y)\mapsto f(x)g(y) $$
is continuous? I somehow struggle with this more than i should... My first idea would be using something like nets:
We have $(x_i,y_i)\to(x,y) \text{ [in $X\times Y$]}\Leftrightarrow x_i \to x \text{[in $X$] and } y_i \to y\text{ [in $Y$]}$ for a net $(x_i,y_i)_{i\in I}$ in the product topology. Therefore it's just $$ \lim_{(x_i,y_i)\to (x,y)} f(x_i)g(y_i) = \lim_{\substack{x_i\to x\\ y_i \to y}} f(x_i)g(y_i) = f(x)g(y) $$ but this feels too sloppy and not very enlightening. Any comments or different approaches would be great.
The map$$\begin{array}{ccc}X\times Y&\longrightarrow&\Bbb R^2\\(x,y)&\mapsto&\bigl(f(x),g(y)\bigr)\end{array}$$is continuous, and so is the map$$\begin{array}{ccc}\Bbb R^2&\longrightarrow&\Bbb R\\(x,y)&\mapsto&xy.\end{array}$$So, you can express your function as the composition of two continuous functions, and that proves that it is continuous.