3 edited body

Here is a sketch of how the proof might go.

If $f:[0,1]\rightarrow\mathbb R$ is continuous but not bounded then the sets $S_n=[0,1]\setminus f^{-1}[(-n,n)]$ are closed with $S_n\ne\emptyset$ for all $n\in\mathbb N$. According to the definition of continuous function the sets $f^{-1}[(-n,n)]$ are represented as $\Sigma_1$ sets of (endpoints of) rational intervals ($\Sigma_1$ definable in the model). So the closed dyadic rational intervals contained in intersecting $S_n$ for $n\in\mathbb N$ can be represented as an infinite $\Pi_1$ (or by a standard trick, equivalently $\Delta_1$) tree, which by Weak König's Lemma must have a path, so $\cap_n S_n\ne\emptyset$ which is absurd.

2 deleted 30 characters in body; added 47 characters in body; edited body

Here is a sketch of how the proof might go.

If $f:[0,1]\rightarrow\mathbb R$ is continuous but not bounded then the sets $S_n=[0,1]\setminus f^{-1}[(-n,n)]$ are closed with $\cap_n S_n=\emptyset$, but $\cap_{n\le N}S_n\ne\emptyset$ S_n\ne\emptyset$for all$N\in\mathbb n\in\mathbb N$. According to the definition of continuous function the sets$f^{-1}[(-n,n)]$are represented as$\Sigma_1$sets of (endpoints of) rational intervals ($\Sigma_1$definable in the model). So the closed dyadic rational intervals contained in$\cap_{n\le N} S_n$for$n\in\mathbb N$can be represented as an infinite$\Delta_1$\Pi_1$ (or by a standard trick, equivalently $\Delta_1$) tree, which by Weak König's Lemma must have a path, so $\cap_n S_n\ne\emptyset$ after allwhich is absurd.

1

Here is a sketch of how the proof might go.

If $f:[0,1]\rightarrow\mathbb R$ is continuous but not bounded then the sets $S_n=[0,1]\setminus f^{-1}[(-n,n)]$ are closed with $\cap_n S_n=\emptyset$, but $\cap_{n\le N}S_n\ne\emptyset$ for all $N\in\mathbb N$. According to the definition of continuous function the sets $f^{-1}[(-n,n)]$ are represented as $\Sigma_1$ sets of (endpoints of) rational intervals ($\Sigma_1$ definable in the model). So the closed dyadic rational intervals contained in $\cap_{n\le N} S_n$ can be represented as an infinite $\Delta_1$ tree, which by Weak König's Lemma must have a path, so $\cap_n S_n\ne\emptyset$ after all.