> [!Theorem] > Let $(f_{n})_{n\in\mathbb{N}}:[0,1]\to \mathbb{R}$ be a sequence of [[Continuous Real Function|continuous function]] which decrease pointwise to $0$. For each $\varepsilon>0$ there is an $n$ for which $f_{n}(x)<\varepsilon$ for all $x\in[0,1]$. > *Proof*. 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 $S$. *Proof*. Suppose the statement were false. Then there is some $\varepsilon>0$ with the property that for every $n$ there is a point $x\in[0,1]$ where $f_{n}(x)\geq\varepsilon$. Pick such a point and call it $x_{n}$. Now if $m > n$ then since the functions decrease at $x_{m}$ we have $f_{n}(x_{m})\geq f_{m}(x_{m})\geq\varepsilon$. By [[Bolzano-Weierstrass Theorem (Sequential Compactness of The Reals)|Bolzano-Weierstrass]], we can find a convergent subsequence of the $(x_{n})$. Let’s say $x_{n_{k}}\to x\in[0,1].$ For each $n$ we have $f_{n}(x_{n_{k}})\geq\varepsilon$ as soon as $n_{k}>n$ and hence $f_{n}(x)\geq\varepsilon$. But this contradicts the fact that $f_{n}(x)\to_{0}$.