# The difference between the statements for sequences of function $f_n(x)$

Let I be an interval and c ∈ I.

Statement A: For all $$\epsilon$$ > 0, there is $$\delta$$ > 0 such that,for all $$n ∈ \mathbb{N}$$ and for all $$x ∈ I$$ satisfying $$|x−c|≤\delta$$, $$|f_n(x)−f_n(c)| ≤ \epsilon$$.

Statement B: For all $$n ∈ \mathbb{N}$$ and for all $$\epsilon > 0$$, there exists $$\delta > 0$$ such that whenever $$x ∈ I$$ and $$|x−c|≤\delta$$, then $$|f_n(x)−f_n(c)| ≤ \epsilon$$.

In my opinion, the difference between the statement is I think statement A says that all the functions are continuous at a certain point, whereas statement B says that each function is continuous at all points

My group says my answer is wrong but I am not sure why.

A: For all $$\varepsilon>0$$, there is a $$\delta>0$$ such that for all $$n\in \Bbb N$$ and [...]
B: For all $$n\in \Bbb N$$ and for all $$\varepsilon>0$$, there is a $$\delta>0$$ such that [...]
The difference here is that in statement A, whatever $$\delta$$ you choose should work for all $$n$$ simulatneously. The constant $$c$$ is still fixed in both cases, so we are only interested in continuity at $$c$$, and not the whole interval.
As an example to illustrate the difference, consider the family of functions $$f_n(x) = nx$$ (and, if you'd like to specify it, $$c = 0$$). It satisfies B, since you can pick $$\delta = \frac\varepsilon n$$. But it does not satisfy A, because there is no single $$\delta$$ you can choose that works for all $$n$$. Higher $$n$$ demands ever smaller $$\delta$$, and you're not allowed to pick $$0$$.
Statement B says each function, by itself, is continuous at $$c$$. It does not in any way state any kind of relationship between the different functions. Statement A says more. It says that not only are they continuous at $$c$$, but also that as $$n$$ grows, their "steepness" (in the sense of $$\varepsilon$$-$$\delta$$, not in the sense of derivative) is bounded.