Restriction of a locally finitely supported function on an ordinal is finitely supported?

Question

This is a question about set theory. Let $f:\delta\rightarrow H$ be a function from an ordinal $\delta< \omega_1$ to an arbitrary abelian group $H$. Endow $\delta$ with the order topology. Let $f$ be a locally finitely supported function, which means that for any $x\in\delta$, there exists an open neighbourhood $U$ of $x$ in $\delta$ such that there are only finitely many points $p_1,\cdots,p_n$ in $U$ such that the value of $f$ on the point is non-zero. My question: is it correct that for any ordinal $\alpha<\delta$, the restriction of $f$ on $\alpha$ is a finitely supported function, i.e. there are only finitely many points in $\alpha$ such that the value of $f$ on the point is non-zero? Thanks in advance for any suggestions.

Answer 1

If the function has infinitely many nonzero values below some ordinal $\alpha$, let $\beta$ be the supremum of the locations of the first $\omega$ many instances. So $\beta\leq\alpha$ and every neighborhood of $\beta$ has infinitely many nonzero values, contrary to assumption.

Answer 2

Yes, this follows from the fact that order topology on any successor ordinal is compact. Let's see a quick proof of this:

Fix a successor ordinal $\alpha+1$ and an open cover $(U_i)_{i \in I}$. Assume for the sake of contradiction that this cover has no finite subcover. Let $\beta \leq \alpha$ be the first element of $\alpha$ such that for every finite $I_0 \subseteq I$, $\bigcup_{i \in I_0} U_i$ fails to cover $[0,\beta]$. Since $\beta < \alpha + 1$, there must be some $j \in I$ such that $\beta \in U_j$.

Now, note that $\beta$ cannot be a successor ordinal. If $\beta = \gamma + 1$, then we'd have that there is a finite $I_0 \subseteq I$ such that $\bigcup_{i \in I_0} U_i$ cover $[0,\gamma]$, implying that we just need to add $U_j$ to this to get a finite subcover of $[0,\gamma]$. Therefore $\beta$ must be a limit ordinal. By the definition of the order topology, there must be some $\gamma < \beta$ such that $[0,\beta] \subseteq [0,\gamma] \cup U_j$. By our choice of $\beta$, there must be a finite $I_0 \subseteq I$ such that $\bigcup_{i \in I_0} U_i$ cover $[0,\gamma]$, but again this implies that $\bigcup_{i \in I_0 \cup \{j\}} U_i$ covers $[0,\beta]$, which is a contradiction.

Therefore no such $\beta$ can exist. $\square$

Now in the specific case of your question, given any $\alpha < \delta$, we can apply the above statement to $\alpha+1$ to get that there is a finite set $\mathcal{U}$ of open sets such that $f$ is finitely supported on each $U \in \mathcal{U}$ and $\mathcal{U}$ covers $[0,\alpha+1]$, which a fortiori implies the same of $[0,\alpha)$.

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One thing to note is that the compactness of $[0,\alpha+1]$ isn't really a special property of ordinals. Any complete linear order is compact in the order topology.