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Quiz 7
Prove that if $f_n \rightrightarrows f$, then $f_n \rightarrow f$.
Proof: Assume that $f_n \colon [a,b] \rightarrow \mathbb{R}$ be a sequence of functions, assume $f \colon [a,b] \rightarrow \mathbb{R}$, and assume that $f_n \rightrightarrows f$ ($f_n$ converges uniformly to $f$), i.e. $\forall \epsilon > 0 \exists N \in \mathbb{N}$ so that if $n \geq N$, $\forall x \in [a,b]$, $|f_n(x)-f(x)|<\epsilon$.

Goal: Prove that $f_n \rightarrow f$ ($f_n$ converges pointwise to $f$), i.e. $\epsilon > 0 \forall x \in [a,b] \exists N \in \mathbb{N}$ so that if $n \geq N$, $|f_n(x)-f(x)|<\epsilon$.

Let $\epsilon>0$ and let $x \in [a,b]$. Since $f_n \rightrightarrows f$, choose $N \in \mathbb{N}$ so that for any $n \geq N$, $|f_n(x)-f(x)|<\epsilon$. This is precisely what it means for $f_n \rightarrow f$, completing the proof. $\blacksquare$