I'm reading a paper where the author makes the following remark:
If $K\subseteq 2^\omega$ is compact and $\phi:K\rightarrow\omega^\omega$ is pseudo-continuous, then $\phi''K$ is bounded.
Let me put some of context: We say that a partial function $\phi:2^\omega\rightarrow\omega^\omega$ is pseudo-continuous iff $f_m\rightarrow f$ is a converging sequence in $dom(\phi)$ then for all $n$: $\forall^\infty_m(\phi(f_m(n))\leq \phi(f(n)))$. As the paper is about characteristic cardinals of the continuum, I think that the word "bounded" in the remark makes reference to the well-knonw order $<^*$.
I tried to get a proof for that remark using that $K$ is sequentally compact (because $2^\omega$ is a metric space) but I didn't obtain nothing.
Please, someone can give me a hint?
Let $\phi \colon K \to \omega^\omega$ be pseudo-continuous where $K$ is a compact subset of $2^\omega$.
Claim. For any $n \in \omega$ and any $x \in K$ there is an open neighborhood $O$ of $x$ and $m \in \omega$ so that for all $y \in O$, $\phi(y)(n) \leq m$.
Proof. Suppose not, as witnessed by $x \in K$, $n \in \omega$. Then we have that for any $k \in \omega$ there is $y_k \in [x \restriction k] \cap K$ so that $\phi(y_k)(n)$ is as large as you wish, say $\phi(y_k)(n) > k$. Obviously $y_k \rightarrow x$, but $\phi(y_k)(n) \rightarrow \infty$ so actually $\forall^\infty k \in \omega (\phi(y_k)(n) > \phi(x)(n))$, contradicting pseudo-continuouity.
Now for fixed $n$, find such an open neighborhood $O_x$ and $m_x$ as above for each $x \in K$. The $O_x$ cover $K$, thus there is a finite set $\{x_0, \dots, x_l \}$ such that $\bigcup_{i \leq l} O_{x_i} = K$. But then we have that for any $y \in K$, $\phi(y)(n) \leq \max \{m_{x_i} : i \leq l \}$.
Now unfix $n$ and get the desired result.