Solutions to [[HW I.pdf]]. 8. Let $S=\{ c\in[0,1]: \forall \varepsilon>0, \exists N\in\mathbb{N}, \forall n \geq N, \forall x\in[0,c], f_{n}<\varepsilon \}$. Since $f_{n}(0)\to 0$ and $f_{n}\geq0$, for all $\varepsilon>0$, there exists $N\in\mathbb{N}$ such that $\forall n\geq N$, $f_{n}(0)<\varepsilon$so $0\in S_{\varepsilon}$. Therefore, by LUBA, $\exists r:=\sup S$. Suppose $r<1$. Then $\forall\varepsilon>0, \exists N\in\mathbb{N}, \forall n\geq N,\forall x\in[0,r],f_{n}(x)<\epsilon$. Continuity gives $\exists \delta>0,|x-r|<\delta \implies |f_{n}(x)-f_{n}(r)|<\varepsilon$. In particular $\forall x\in\left[ r,r+\frac{\delta}{2} \right], f_{n}(r)-f_{n}(x)>-\varepsilon\implies f_{n}(x)<f_{n}(r) + \varepsilon < 2\varepsilon$ for all $n \geq N$. So $r+\frac{\delta}{2}\in S$ which contradicts the fact that $r$ is upper bound for $S$.