[proofplan] The proof is a direct unpacking of the definitions. Uniform boundedness gives one constant $M$ that bounds every function $f_n$ at every point of $E$. A subsequence only selects terms whose indices $n_k$ are still natural numbers, so the same inequality applies to each selected term without changing the constant. [/proofplan] [step:Choose the common bound for the original sequence] By the uniform boundedness hypothesis, there exists a constant $M \geq 0$ such that \begin{align*} |f_n(x)| \leq M \end{align*} for every $n \in \mathbb{N}$ and every $x \in E$. [/step] [step:Apply the original bound to each selected index] Let $(n_k)_{k=1}^{\infty}$ be a strictly increasing sequence in $\mathbb{N}$. For each $k \in \mathbb{N}$, the number $n_k$ is an element of $\mathbb{N}$, so the bound from the original sequence applies with $n = n_k$. Hence, for every $k \in \mathbb{N}$ and every $x \in E$, \begin{align*} |f_{n_k}(x)| \leq M. \end{align*} [guided] We must show that the subsequence has one bound that works simultaneously for all subsequence terms and all points of $E$. The original hypothesis already gives a constant $M \geq 0$ such that \begin{align*} |f_n(x)| \leq M \end{align*} whenever $n \in \mathbb{N}$ and $x \in E$. Now fix an arbitrary subsequence $(f_{n_k})_{k=1}^{\infty}$. By definition of a subsequence, $(n_k)_{k=1}^{\infty}$ is a strictly increasing sequence of natural numbers. In particular, for each fixed $k \in \mathbb{N}$, the index $n_k$ is one of the admissible indices in the original bound. Therefore we may substitute $n_k$ for $n$ in the inequality above. This gives \begin{align*} |f_{n_k}(x)| \leq M \end{align*} for every $k \in \mathbb{N}$ and every $x \in E$. The important point is that the constant has not changed: the same $M$ that bounded the entire original sequence also bounds the selected terms, because no subsequence term lies outside the original sequence. [/guided] [/step] [step:Conclude uniform boundedness of the subsequence with the same bound] The preceding inequality is exactly the definition of uniform boundedness on $E$ for the sequence $(f_{n_k})_{k=1}^{\infty}$, with common bound $M$. Since the subsequence was arbitrary, every subsequence of $(f_n)_{n=1}^{\infty}$ is uniformly bounded on $E$ with the same common bound. [/step]