about continuous functions Let $f,g: \mathbb R \to \mathbb R$ be continuous. How may I show that the following are also continuous functions


*

*$f+g$

*$fg$   

*$f\circ g$

 A: To go against the current: if you use the definition of continuous from topology,

Let $X$ and $Y$ be topological spaces. A function $f\colon X\to Y$ is continuous if and only if for every open subset $\mathscr{O}\subseteq Y$, $f^{-1}(\mathscr{O})$ is an open subset of $X$.

and you think of $\mathbb{R}$ as having the usual topology (so open sets are arbitrary unions of open intervals), then the proof of (iii) is very easy: let $\mathscr{O}$ be an open subset of $\mathbb{R}$. Then $(f\circ g)^{-1}(\mathscr{O}) = g^{-1}(f^{-1}(\mathscr{O}))$; but $f$ is continuous, so $f^{-1}(\mathscr{O})$ is open, and $g$ is continuous, hence $g^{-1}(f^{-1}(\mathscr{O}))$ is also open. So $f\circ g$ is continuous.
(i) and (ii), though, are a bit more complicated with this definition, though, so you probably want to do $\epsilon$-$\delta$ proofs for those.
A: "$f+g$":
Let $x\in\mathbb R$ and $\epsilon > 0$.


*

*By continuity of $f$ at $x$ there is $\delta_1>0$ such that for every $y\in\mathbb R$ with $|x-y|<\delta_1$ we have $|f(x)-f(y)|< \epsilon / 2$.

*By continuity of $g$ at $x$ there is $\delta_2>0$ such that for every $y\in\mathbb R$ with $|x-y|<\delta_2$ we have $|g(x)-g(y)|< \epsilon / 2$.

*Let $\delta = min\{\delta_1, \delta_2\}$. For every $y \in \mathbb R$ with $|x-y|<\delta$ we have $|f(x) + g(x) - f(y) - g(y)| \leq |f(x) - f(y)| + |g(x) - g(y)| < \epsilon/2 + \epsilon/2 = \epsilon$.
Now try the other two as they are similar.
A: If you use this definition of continuity:
(DEF) $f:X \rightarrow Y$ is continuous at $x_0$ iff $\forall \varepsilon > 0 \exists \delta > 0 : |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon$
Then for the case (i) you proceed as follows:
you want to show that $f + g$ is continuous at $x_0$. You know, because  $f$ and $g$ are continuous that for each $f$ and $g$ respectively you can find a $\delta_f$ and a $\delta_g$ such that 
$$|x - x_0| < \delta_f \implies |f(x) - f(x_0)| < \frac{\varepsilon}{2} $$ and  $$|x - x_0| < \delta_g \implies |g(x) - f(x_0)| < \frac{\varepsilon}{2} $$ for any given $\varepsilon$.
Now you pick $\delta := min(\delta_f, \delta_g)$ and then you use the triangle inequality as follows:
$|(f+g)(x) - (f+g)(x_0)| = |f(x) + g(x) - f(x_0) -g(x_0)| \leq |f(x) - f(x_0)| + |g(x) - g(x_0| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$
where the last inequality follows from how you chose $\delta$.
The cases ii) and iii) are similar. 
A: Use general principles:
(a) The basic operations in ${\mathbb R}$ resp. ${\mathbb C}$ are continuous where defined.
(b) A vector valued function $f=(f_1,f_2)$ is continuous iff the $f_i$ are continuous.
(c) The composition of two continuous functions is continuous.
Now $f\cdot g= p \circ h$ where $h: x\mapsto\bigl(f(x),g(x)\bigr)$
and  $p$ denotes the product function.
A: use the epsilon delta definition
so you want to show f+g is continuous at x. so take a delta for f+g. that delta gives you an epsilon for f and another epsilon for g (by continuity). then you need to come up with an epsilon for f+g.
