Let $X$ be a topological space and let $f:X \to \mathbb{R}$ ,$g:X \to \mathbb{R}$ be continuous functions. Show that $fg$ is continuous.
My work:
To show $fg$ is continuous at $x$ for each $x \in X$, let $y=fg(x)$.
To show if $N_y$ is a neighborhood of $y$, then the pre-image of $y$ is a neighborhood of $x$.
I know that there exists $B_\epsilon(y) \in N_y$ so I want to show that $（fg）^{-1}(B_\epsilon(y)) \in N(x)$
Let $N_x=（fg）^{-1}(B_\epsilon(y))$, I want to find an open set in $N_x$.
Can anyone give me a hint of how to choose such open set or idea how to prove this ?
Best Answer
Since the topology in ${\bf{R}}$ is still the standard Euclidean, it is still a matter of proving this in the classical way that $|f(x)g(x)-f(x_{0})g(x_{0})|\leq|f(x)-f(x_{0})||g(x)|+|f(x_{0})||g(x)-g(x_{0})|$. Now the boundedness of $|g(x)|$ around $x_{0}$ is given by the continuity of $g$ at $x_{0}$ as well. Note that one could see that $|x-x_{0}|<\delta$ is replaced by some neighbourhood $\mathcal{N}(x_{0})$, and the matter of $\min\{\delta,\delta'\}$ should be replaced by the intersection of two neighbourhoods.