$R^{n}$ Topology: Open sets Let $f,g : X \rightarrow \mathbb{R}$ be continuous functions in $X \subset \mathbb{R}^{n}$. 
Why the set $A = \{ x\in X; f(x) < g(x) \}$ is open in $X$?
Okay, i tried to find a open set $ U \subset \mathbb{R}^{n}$ such that $A = U \cup X$ but o dont know when use the hyphotesis that $f(x) < g(x)$. Any help?
 A: $g-f$ is continuous, and your set is $(g-f)^{-1}(0,+\infty)$.
A: Hint: prove that the complement is closed. Let $(a_n)$ a sequence in $X$ such that $a_n\to a$, and use that $f(a_n)\to f(a)$ (and with $g$)
A: This is because if $f$ and $g$ are continuous, so is $f - g$.  
Now, $\{x \in X : f(x) < g(x) \} = \{x \in X : f(x) - g(x) < 0 \}$.  So our set is the set of $x \in X$ such that $f-g$ evaluated at those $x$ is less than $0$.  In other words, we want to find the $x \in X$ so that $f - g$ takes those $x$ into $(-\infty, 0)$.  The notation for this set is $(f-g)^{-1}( (-\infty, 0))$.  
Since $(-\infty, 0)$ is an open set, and $f- g$ is continuous, by definition of continuity $(f - g)^{-1}(-\infty, 0)$ is open.  Since this equals our set $\{x \in X : f(x) < g(x) \}$, that means this set is open.
A: I'm guessing you haven't covered the result that a function $f\colon\mathbb R^n\to\mathbb R^m$ is continuous (in the $\epsilon$-$\delta$ sense) if and only if the preimage $f^{-1}(U)$ of every open set $U\subset\mathbb R^m$ is open, because that would make the question a triviality (as in the other answers).  
In fact, it's not difficult to prove this directly.  I think you can do it yourself, so I'm just going to remind you of the steps you need to go through.  


*

*A set $U\subset \mathbb R^m$ is open if whenever $x\in U$ there exists $\delta>0$ such that $B_x(\delta)\subset U$, where
$$
B_x(\delta)=\{y\in\mathbb R^n\;\colon\;\|x-y\|<\delta\}
$$
In other words, for all $x\in U$, we require that there exists $\delta>0$ such that $y\in U$ whenever $\|x-y\|<\delta$.  

*OK, so we want to show that this holds for $\{x\in X\;\colon\;f(x)<g(x)\}$.  So let $x$ be an element of this set; i.e., let $x$ be such that $f(x)<g(x)$.  

*Now we need to show that there exists some $\delta>0$ such that $f(y)<g(y)$ whenever $\|x-y\|<\delta$.  

*In order to do this, we use the fact that $f$ and $g$ are continuous at $x$: so for any $\epsilon>0$ there exists some $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ for all $y$ such that $\|x-y\|<\delta$.  How can we use this to find a $\delta$ as above?  What value shoudl we take for $\epsilon$.  
You might find this slightly easier if you use the fact that the function $f-g$ is continuous (as other answers have pointed out), but it isn't necessary.
I hope that's enough to be getting along with.
A: Try looking at the function $g(x) - f(x)$, which is also continuous.
