[proofplan] We fix an arbitrary point $x\in X$ and reduce the assertion to the scalar sequence $(f_k(x))_{k=1}^{\infty}$ in $\mathbb{R}$ or $\mathbb{C}$. Pointwise convergence gives convergence of this scalar sequence, and a direct $\varepsilon$-argument shows that every convergent scalar sequence is bounded. The resulting bound may depend on $x$, which is exactly the requirement in pointwise boundedness. [/proofplan]

[step:Fix a point and name the pointwise scalar limit] Let $x\in X$ be arbitrary. Since $(f_k)_{k=1}^{\infty}$ converges pointwise on $X$, there exists a function $f:X\to\mathbb{F}$ such that, for this fixed $x$, \begin{align*} \lim_{k\to\infty} f_k(x) = f(x). \end{align*} Define the scalar sequence $a:\mathbb{N}\to\mathbb{F}$ by $a_k:=f_k(x)$ for every $k\in\mathbb{N}$, and define the scalar limit $a\in\mathbb{F}$ by $a:=f(x)$. Then $(a_k)_{k=1}^{\infty}$ converges to $a$ in $\mathbb{F}$. [/step]

[step:Bound the convergent scalar sequence at the fixed point]Because $a_k\to a$, applying the definition of convergence with $\varepsilon=1$ gives an index $N\in\mathbb{N}$ such that \begin{align*} |a_k-a|<1 \end{align*} for every $k\ge N$. Hence, by the triangle inequality in $\mathbb{R}$ or $\mathbb{C}$, \begin{align*} |a_k|\le |a|+|a_k-a|<|a|+1 \end{align*} for every $k\ge N$. Define the finite set of [real numbers](/page/Real%20Numbers) \begin{align*} S_x:=\{|a_1|,\dots,|a_{N-1}|,|a|+1\}. \end{align*} If $N=1$, interpret the initial list $|a_1|,\dots,|a_{N-1}|$ as empty, so that $S_x=\{|a|+1\}$. Since $S_x$ is finite and nonempty, it has a maximum. Define \begin{align*} M_x:=\max S_x. \end{align*} Then $M_x<\infty$. If $1\le k<N$, then $|a_k|\le M_x$ by the definition of $M_x$. If $k\ge N$, then $|a_k|<|a|+1\le M_x$. Therefore \begin{align*} |a_k|\le M_x \end{align*} for every $k\in\mathbb{N}$.[/step]

[guided]We now prove the boundedness of the scalar sequence directly, rather than citing it as a separate theorem. The convergence $a_k\to a$ means that every prescribed error tolerance is eventually achieved. We choose the specific tolerance $\varepsilon=1$ because boundedness only requires some finite tail bound, not a sharp one. Thus there exists an index $N\in\mathbb{N}$ such that \begin{align*} |a_k-a|<1 \end{align*} for every $k\ge N$. For all terms in this tail, the triangle inequality gives \begin{align*} |a_k|\le |a|+|a_k-a|<|a|+1. \end{align*} This controls every term from index $N$ onward. The only terms not yet controlled are the finitely many initial terms $a_1,\dots,a_{N-1}$. To put the initial part and the tail under one bound, define \begin{align*} S_x:=\{|a_1|,\dots,|a_{N-1}|,|a|+1\}. \end{align*} When $N=1$, there are no initial terms before the tail, so this means $S_x=\{|a|+1\}$. In every case $S_x$ is a finite nonempty subset of $\mathbb{R}$, so it has a maximum. Define \begin{align*} M_x:=\max S_x. \end{align*} Then $M_x$ is finite. If $1\le k<N$, the number $|a_k|$ is one of the entries used to form $S_x$, so $|a_k|\le M_x$. If $k\ge N$, the tail estimate gives $|a_k|<|a|+1$, and $|a|+1\le M_x$ because $|a|+1\in S_x$. Hence \begin{align*} |a_k|\le M_x \end{align*} for every $k\in\mathbb{N}$.[/guided]

[step:Translate the scalar bound back to pointwise boundedness] For every $k\in\mathbb{N}$, the definition $a_k:=f_k(x)$ gives \begin{align*} |f_k(x)|=|a_k|\le M_x. \end{align*} The point $x\in X$ was arbitrary, and the constant $M_x$ is allowed to depend on $x$. Therefore, for every $x\in X$ there exists $M_x<\infty$ such that $|f_k(x)|\le M_x$ for every $k\in\mathbb{N}$. This is precisely pointwise boundedness of $(f_k)_{k=1}^{\infty}$ on $X$. [/step]