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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.

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Look at what the difference between the statements is:

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.

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