Something wrong with my proof on little-o Big-O arithmetics I run into this problem in a web site and it states that the equality below is false. When I try to prove that this is false I end up finding out that it is true. What I wanted to ask you is where I did wrong in the proof below. I dont give up quickly but it's been three days, it started really to put me back into my schedule.
Do you think my proof is false in any step?
Thanks in advance!
The question screenshot
$o(x^n) + o(x^m) = O(x^n), {x \to \infty}, n>m$
$ f(x) = o(x^n) \iff |f(x)| < \epsilon * |x^n| , \forall x \geq N $
$ g(x) = o(x^m) \iff |g(x)| < \epsilon * |x^m| , \forall x \geq M $
$ |f(x)| + |g(x)| \leq \epsilon * (|x^n| + |x^m|) , \forall x \geq \max(M,N) $
$ |f(x) + g(x)| \leq |f(x)| + |g(x)| \leq \epsilon * (|x^n| + |x^m|) , \forall x \geq \max(M,N) $ (triangle inequality)
$ |f(x) + g(x)| \leq \epsilon * (|x^n| + |x^m|) , \forall x \geq \max(M,N) $  (logical result of the statement above)
$ |f(x) + g(x)| \leq \epsilon * (|x^n| + |x^m|) \leq \epsilon * (|x^n| + |x^n|) , \\ \forall x \geq \max(M,N)\ \  (\text{if }\ \  n>m\ \text{ then}\ \  x^n > x^m\ \  if\ \ x\ \to \infty) $
$ |f(x) + g(x)| \leq \epsilon * 2 * |x^n|, \\ \forall x \geq \max(M,N) $ (logical result of the statement above)
$ |f(x) + g(x)| \leq \epsilon'  * |x^n|, \forall x \geq \max(M,N),\ (where\ \epsilon' = 2*\epsilon)$
$ f(x) + g(x) = o(x^n), \forall x \geq \max(M,N) $ (formal definition of little-o)
$ o(x^n) + o(x^m) = o(x^n), \forall x \geq \max(M,N) $ (replace f(x) and g(x) back)
$ o(x^n) + o(x^m) = O(x^n), \forall x \geq \max(M,N) $ (little-o implies big-O)
 A: I will assume that the subscripts are both typographical errors (what a lousy test!) and should actually be superscripts. Secondly, as per the comments it uses a common abuse of notation where people write "$f = O(g)$" to mean "$f ∈ O(g)$". I shall just state the precise facts:
Take any $m,n∈ℕ$ such that $m<n$.

*

*As $x → ∞$, we have $o(x^n)+o(x^m) ⊆ o(x^n)+o(x^n) ⊆ o(x^n) ⊆ O(x^n)$, where the 2nd and 3rd steps are obvious and the 1st is because eventually $x > 1$ and so $x^m ≤ x^n$. (I will leave you to try the actual $ε,δ$-proof.)


*As $x → ∞$, we have $o(x^n) ⊆ o(x^n)+o(x^m)$, and so by fact 1 we have $o(x^n)+o(x^m) = o(x^n)$.


*As $x →∞$, we have $o(x^n)+o(x^m) = o(x^n) ≠ O(x^n)$ because $x^n ∈ O(x^n)$ but $x^n ∉ o(x^n)$.
People who abuse notation write $o(x^n)+o(x^m) = O(x^n)$, which is wrong because it contradicts the third fact above.
As for your proof attempt, you neither defined not quantified $ε$, so it is wrong. Look at the actual formal definitions of the asymptotic notations; they do not look like yours.
