Constant continuity theorem? I am proving continuity of functions and have come across a question. In my book and notes I have many theorems but I don't seem to have anything about constants affecting continuity. That is, I'm proving that $8^x$ is continuous. We are to assume $2^x$ is already continuous. Thus I broke $8^x=4\cdot2^x$. I have a theorem where $fg$ is continuous where $f$ and $g$ are real-valued functions. So should I assume $f(x)=4$ as a function or is there a theorem for constants? Or is it fine to just assume that constants would not affect continuity?
 A: $8^x=4 \cdot 2^x$ is not right.
$8^x=(2^x)^3$ is right and useful.
A: (Iterate what others have said about $8^x \neq 4 \cdot 2^x$.)
If you want to use another definition of continuity (that the preimage of every open set is open)...
A constant function takes only one value, $y$, in the codomain.  Let $U$ be any open set in the codomain.  If $y \in U$, the preimage of $U$ is the entire domain, which is open.  If $y \not \in U$, the preimage of $U$ is the empty set, which is open.  Either way, the preimage of every open set is open, so a constant function is continuous.
Added in edit:  Note that we have required nothing about the topologies of the domain or codomain.  Constant functions are continuous regardless of the topologies of these two spaces.
A: $8^x \neq 4 \cdot 2^x$. What is instead true is that $8^x=(2^3)^x= 2^{3x}=(2^x)^3$, which is the product of three continuous functions by the assumption.
(But yes, a constant function is continuous: it is equal to its on limit at every point, for example, which is one characterisation of continuity.)
