A question on the equality between two sets. Let $\{f_n\}_{n\in\mathbb{N}}$ a sequence of function on $X$ to $[-\infty,\infty]$ and let $\alpha\in\mathbb{R}$. I must prove that 

$$ \bigg\{x\in X\;\bigg|\sup_nf_n(x)>\alpha\bigg\}=\bigcup_{n=1}^\infty \big\{x\in X\;|\,f_n(x)>\alpha\big\}$$

Proof. Let $x\in\cup_n\{f_n>\alpha\}$, exists $N\in\mathbb{N}$ such that $f_N(x)>\alpha$. As $\sup_nf_n(x)\ge f_n$ for all $n\in\mathbb{N}$ we have that $\sup_n f_n(x)\ge f_N(x)>\alpha$.
Viceversa, let $x\in\{\sup f_n>\alpha\}$ we have that $\sup_nf_n(x)>\alpha$. Fixed $\varepsilon>0$ exists $N\in\mathbb{N}$ such that $f_N(x)>\sup_n f_n(x)-\varepsilon>\alpha-\varepsilon$, then $f_N(x)>\alpha-\varepsilon$ for all $\varepsilon >0$, therefore $f_N(x)\ge\alpha$, then $x\notin\cup_n\{f_n>\alpha\}$.

Question. Where am I wrong?

Thanks!
 A: There is a small point you need to be more careful about: At the end, the non-strict inequality "$f_N(x)\geq \alpha$" of course does not imply that $x\not\in\left\{f_n>\alpha\right\}$. For example, if $\alpha=2$ and all $f_n$ are constant, $f_n=3$, we always have $f_N(x)\geq\alpha$ and $f_N(x)>\alpha$.
A more precise statement would be that "$f_N(x)\geq \alpha$, however this does not necessarily imply that $f_N(x)>\alpha$".

The second problem is the dependency on $N$ and $\varepsilon$. As you correctly stated, "for every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that [...] $f_N(x)>\alpha-\varepsilon$". But then you let $\epsilon\to 0$ on the right-hand side and kept the left-hand side as it is, even though $N$ depends on $\varepsilon$.
To be more precise, we could state "For every $\varepsilon>0$, there exists $N(\varepsilon)\in\mathbb{N}$ such that [...] $f_{N(\varepsilon)}(x)>\alpha-\varepsilon$". With this (a little cumbersome) notation it is clearer that any sort of "limit on $\varepsilon$" needs to be taken on both sides.

Finally, as for the correct proof: Recall the property of the supremum:

The supremum of a subset $A\subseteq\mathbb{R}$ is the only number $\sup(A)$
  
  
*
  
*$a\leq\sup(A)$ for all $a\in A$;
  
*For every $\epsilon>0$, there exists $a\in A$ with $\sup(A)-\varepsilon<a$.
  

This works fine whenever $\sup(A)$ is finite. A more general description, which also works in the infinite case, and for more general Partially Ordered Spaces is the following. 

The supremum of $A$ is the only number $\sup(A)$ such that
  
  
*
  
*$a\leq\sup(A)$ for all $a\in A$;
  
*For every $b<\sup(A)$, there exists $a\in A$ with $b<a$.
  

If you use the second description of supremums, the exercise should become clearer: A number $\alpha$ is smaller than the supremum $\sup f_n(x)$ if and only if there exists an element of $\left\{f_n(x):n\in\mathbb{N}\right\}$ greater than $\alpha$; in other words, there is $n\in\mathbb{N}$ such that $\alpha<f_n(x)$. This is precisely the description of the right-hand side.
Alternatively, using the second description, note that the inequality defining supremums deals with the supremum itself, and a number smaller than it. These are precisely the ingredients we have!
More precisely, there is $N$ such that $\alpha<f_N(x)\leq\sup f_n(x)$. If you want to use $\varepsilon$s, let $\varepsilon=\sup f_n(x)-\alpha$.
A: You just need to use the fact that $\sup_n f_n > \alpha$ implies the existence of an $N$ such that $f_N > \alpha$, which can be very easily seen by contradiction.
