How does the order of choosing x and epsilon distinguish pointwise convergence from uniform convergence?

Conditions

• A sequence of functions $f_n$ and a limit function $f$ are defined on a region.
• Pointwise convergence: $\forall x, \forall \epsilon>0, \exists N$ such that $n\ge N \implies |f(x)-f_n(x)|<\epsilon$.
• Uniform convergence: $\forall \epsilon>0, \exists N$ such that $n\ge N \implies |f(x)-f_n(x)|<\epsilon$ for all $x$ in the region.

Reasoning, step by step

1. For pointwise convergence, pick a point $x$ first.
2. Then pick a tolerance $\epsilon>0$.
3. Find an index $N$ that satisfies the condition for that specific $x$ and $\epsilon$ (N can depend on x).
4. For uniform convergence, pick a tolerance $\epsilon>0$ first.
5. Find an index $N$ that satisfies the condition for all $x$ in the region simultaneously (N must not depend on x).

Example

The lecturer explicitly notes that the order changes. The slide now begins with “First pick $ε > 0$,” and then asks whether there is an index N such that whenever $n \ge N$, the inequality |$f(x) - f_n(x)$| < ε holds “for all x in a region?” The same algebraic inequality appears, but the quantifier structure is different because x has not been fixed in advance.

Common misconceptions

• Thinking that because both definitions use the same inequality $|f(x)-f_n(x)|<\epsilon$, they mean the same thing.
• Assuming that $N$ in pointwise convergence must also be independent of $x$.

Watch the explanation

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