You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
Since S122 is second countable, its enough to show that $U_i(\gamma)$ and $V_i(n, w)$ are regular. In fact we'll show they are zero-dimensional. For this it suffices to show that $Z = Y\cup \{(n, w)\}$ is zero-dimensional.
Suppose that $\gamma\in \text{cl}_Z V_i(n, w)$. Then $U_j(\gamma)\cap V_i(n, w)\neq\emptyset$ for all $j$. It follows that, say, $U_j(\gamma)\cap U_i(\alpha q(n, w))\neq\emptyset$ for all $j$, $\alpha\in w$. So there exists $\beta_j$ which extends both $\alpha q(n, w)$ and $\gamma$. If $\alpha q(n, w)$ extends $\gamma$, then we must have $q(n, w)\geq j$ for all $j$, which is impossible. So $\gamma$ extends $\alpha q(n, w)$ and since $\beta_j\in U_i(\alpha q(n, w))$, we must have $\gamma\in U_i(\alpha q(n, w))$. So $V_i(n, w)$ is clopen in $Z$.
Similarly, by item #2 for the space #125, we have that $\text{cl}_Z U_i(\gamma) =Z\cap \text{cl}_X(U_i(\gamma)) =Z\cap (U_i(\gamma)\cup Z(i, \gamma)) = U_i(\gamma) \cup (\{(n, w)\}\cap Z(i, \gamma))$. By taking $i$ large enough so that $i > q(n, w)$, we have $(n, w)\notin Z(i, \gamma)$ and so $U_i(\gamma)$ is clopen in $Z$.
It follows that $Z$ is regular, and so $V_i(n, w)$ and $U_i(\gamma)$ are also regular. Thus $X$ is locally metrizable.
Trait Suggestion
The space Gustin's sequence space S122 is Locally metrizable P82, but this fact is not known to pi-Base today:
link to pi-Base
Proof/References
Let$U_i(\gamma)$ and $V_i(n, w)$ be as in the definition of space #125 in DOI 10.1007/978-1-4612-6290-9_6.
Since S122 is second countable, its enough to show that$U_i(\gamma)$ and $V_i(n, w)$ are regular. In fact we'll show they are zero-dimensional. For this it suffices to show that $Z = Y\cup \{(n, w)\}$ is zero-dimensional.
Suppose that$\gamma\in \text{cl}_Z V_i(n, w)$ . Then $U_j(\gamma)\cap V_i(n, w)\neq\emptyset$ for all $j$ . It follows that, say, $U_j(\gamma)\cap U_i(\alpha q(n, w))\neq\emptyset$ for all $j$ , $\alpha\in w$ . So there exists $\beta_j$ which extends both $\alpha q(n, w)$ and $\gamma$ . If $\alpha q(n, w)$ extends $\gamma$ , then we must have $q(n, w)\geq j$ for all $j$ , which is impossible. So $\gamma$ extends $\alpha q(n, w)$ and since $\beta_j\in U_i(\alpha q(n, w))$ , we must have $\gamma\in U_i(\alpha q(n, w))$ . So $V_i(n, w)$ is clopen in $Z$ .
Similarly, by item #2 for the space #125, we have that$\text{cl}_Z U_i(\gamma) =Z\cap \text{cl}_X(U_i(\gamma)) =Z\cap (U_i(\gamma)\cup Z(i, \gamma)) = U_i(\gamma) \cup (\{(n, w)\}\cap Z(i, \gamma))$ . By taking $i$ large enough so that $i > q(n, w)$ , we have $(n, w)\notin Z(i, \gamma)$ and so $U_i(\gamma)$ is clopen in $Z$ .
It follows that$Z$ is regular, and so $V_i(n, w)$ and $U_i(\gamma)$ are also regular. Thus $X$ is locally metrizable.
Comment
This also resolves the property Locally pseudometrizable P144
The text was updated successfully, but these errors were encountered: