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**anon2194** $\displaystyle f,g : A \rightarrow R$ are continuous on A.

We are given that:

$\displaystyle \forall c \in A, \forall \epsilon > 0, \exists d_{f}(\epsilon, c): \forall x \in A, 0 < |x - c| < d_{f} \rightarrow |f(x) - f(c)| < \epsilon$

$\displaystyle \forall c \in A, \forall \epsilon > 0, \exists d_{g}(\epsilon, c): \forall x \in A, 0 < |x - c| < d_{g} \rightarrow |g(x) - g(c)| < \epsilon$

$\displaystyle \forall x,c \in A$,

$\displaystyle |fg(x)-fg(c)| = |fg(x)-f(x)g(c)+f(x)g(c)-fg(c)|$

$\displaystyle |fg(x)-fg(c)| \le |f(x)|*|g(x)-g(c)|+|g(c)|*|f(x)-f(c)|$

Let $\displaystyle M(x,c)=$max{$\displaystyle f(x), g(c)$}

Then $\displaystyle |fg(x)-fg(c)| \le M(x,c)*(2\epsilon)$

i.e. $\displaystyle \forall c,x \in A, \forall \epsilon_{1}:=\frac{\epsilon}{2M(x,c)}>0, \exists \delta_{fg}=$min{$\displaystyle d_f(\epsilon,c),d_g(\epsilon,c)$}$\displaystyle :0<|x-c|<\delta(\epsilon,c) \rightarrow |fg(x)-fg(c)|<\epsilon_{1}$

Then fg is continuous on A.

This is my proof for the stated theorem. Is there any way I can improve it?