Prove rigorously in arbitrary metric spaces that $$\lim_{x \to a} \phi (x) + \gamma (x) = \lim_{x \to a} \phi (x) + \lim_{x \to a} \gamma(x)$$
This is what I tried to do before getting stuck (take note, what I tried to do, may be completely wrong, in which case, please let me know)
My Attempted Proof
Let $\alpha, \phi, \gamma$ be mappings on $E \subset X$, to $Y$, where $X$ and $Y$ are metric spaces.
Put $\alpha (x) = \phi (x) + \gamma (x)$. Then $\lim_{x \to a} \phi (x) + \gamma (x) = \lim_{x \to a} \alpha (x)$
Suppose, $\lim_{x \to a} \alpha (x)$ exists, then by the definition of $\lim$, there eixsts a if there is a $q \in Y$ such that $\forall \epsilon > 0, \exists \delta > 0$ such that
$$d_Y (\alpha(x), q) = d_Y \left([\phi (x) + \gamma (x)], q\right) < \epsilon \ \ \ \ \ \ \ \ \ \ \ (1)$$
for all points $x \in E$ for which $0 < d_x(x,a) < \delta$.
Likewise $\lim_{x \to a} \phi (x) = c$ is defined if there exits a $c$ such that $\forall \epsilon > 0, \exists \delta > 0$ such that $$d_Y (\phi(x), c) < \frac{\epsilon}{2} \ \ \ \text{for all points $x \in E$, for which} \ \ \ 0 < d_x(x, a) < \delta \ \ \ \ \ \ (2)$$
Again $\lim_{x \to a} \gamma (x) = d$ is defined if there exits a $c$ such that $\forall \epsilon > 0, \exists \delta > 0$ such that $$d_Y (\gamma(x), d) < \frac{\epsilon}{2} \ \ \ \text{for all points $x \in E$, for which} \ \ \ 0 < d_x(x, a) < \delta \ \ \ \ \ \ (3)$$
Assuming $c$ and $d$ exist we have $$d_Y (\gamma(x), d) + d_Y (\phi(x), c) < \epsilon$$
After this point I'm lost as to where to go further.
I need to show the equality of the limits (which I'm not sure entirely how to do), and the fact that the $\delta$ chosen in my initial assumption in $(1)$, works for $(2)$ and $(3)$. Furthermore I've made an additional assumption that $c$ and $d$ must exist, when their existence needs to follow from $(1)$.
Right now it seems as if I have 3 seemingly disjointed statements and no real way to connect them into a logical proof. How would I complete this proof?
I think there is something you need to think about for this proof. The precise proof you are looking for is:
If $\lim_{x \rightarrow a} \phi(x)=c$ and $\lim_{x \rightarrow a} \gamma(x)=d$, then $\lim_{x \rightarrow a} \phi(x)+\gamma(x)=c+d$.
I am going to assume the space is endowed with a norm, which means $d(x,y)=||x-y||$.
Given $\epsilon>0$, you can find $\delta_{1}$ such:
$d_{Y}(\phi(x),c)<\epsilon/2$, $\forall x$, for which $0<d(x,a)<\delta_{1}$
And $\delta_{2}$ such:
$d_{Y}(\gamma(x),d)<\epsilon/2$, $\forall x$, for which $0<d(x,a)<\delta_{2}$
Then, given $\epsilon>0$, there is $\delta:=\min\{\delta_{1},\delta_{2}\}$ satisfying ($\forall x$, for which $0<d(x,a)<\delta$):
$d_{Y}(\phi(x)+\gamma(x),c+d)=||(\phi(x)+\gamma(x))-(c+d)||=||(\phi(x)-c)-(\gamma(x)-d)|| \leq ||(\phi(x)-c)|| + ||(\gamma(x)-d)||=d_{Y}(\phi(x),c)+d_{Y}(\gamma(x),d)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
And this is what we can prove (I have used triangular inequality).
We can't prove the statement in the other way, as it is false. For example, for $\phi(x)=\frac{1}{x}$, $\gamma(x)=\frac{-1}{x}$ and $a=0$. In this case, $\lim_{x \rightarrow 0} \frac{1}{x}-\frac{1}{x}=0$, but $\lim_{x \rightarrow 0} \frac{1}{x}=\lim_{x \rightarrow 0} \frac{-1}{x}=\infty$.