If $f$ and $g$ are continuous maps on a topological space $(X,\tau)$ then $f+g$ is 
If $f$ and $g$ are continuous maps on a topological space $(X,\tau)$ then $f+g$ is
$(a)$ maybe continuous.
$(b)$ may or may not be continuous.
$(c)$ is continuous.
$(d)$ none of these.

I think the answer should be continuous which is option $(c),$ but my book says option $(a)$ i.e. maybe continuous. I don't understand how do I prove this or contradict this. Any help is appreciated thanks. An example demonstrating the fact that if $f,g$ are continuous then $f+g$ may be continuous will also work. Thanks.
 A: Start by thinking about what does $f+g$ really mean. There is a hidden composition of two functions here. Define $h:X\to \mathbb R^2$ (or change $\mathbb R$ for wherever space you are summing) by $h(x) = (f(x),g(x))$ and define $(+): \mathbb R^2 \to \mathbb R$ as $(+)(x,y) = x+y$ (this is essentially what's called polish notation). Now, you have $f+g = (+) \circ h$. In other words, you first compute $f(x), g(x)$ and then you add them.
Now you have split a hard problem into three easier problems:

*

*Is $h$ continuous? (Remember that $f$, $g$ are continuous)

*Is $(+)$ continuous?

*Is the composition of two continuous functions (namely $(+)$ and $h$)  continuous?

A: In a general topological space, what does $+$ even mean?
If we assume that $f$ and $g$ are maps into euclidean space, say, then for any point $x_0$, $\forall\epsilon$ we need a $\delta\gt0$, such that we have $|x-x_0|\lt\delta\implies |(f+g)(x)-(f+g)(x_0)|\lt\epsilon$.  Just choose $\delta$ such that $|x-x_0|\lt\delta\implies |f(x)-f(x_0)|\lt\epsilon/2$ and $|g(x)-g(x_0)|\lt\epsilon/2$.
It is easy to see that this $\delta$ will do the trick, given that $(f+g)(x):=f(x)+g(x)$.
