pi-base
Completing Line with two origins (S83)
@prabau I essentially just copied your proof on mathse and adjusted it for Strongly Paracompact. Not sure if theres a better way.
P145: Please let's try to be careful. There is an annoying, awkward turn of phrase here. In general, when one says something like "... some $a\in A$ ...", "some" qualifies the immediately following $a$ and not $A$. So it means some element $a$ such that $a$ is contained in $A$ and does not mean some $A$ such that $A$ contains $a$.
Also, $(U\cap U_0)\setminus{0_1}$ ?
And what is $U \in \mathcal{V}, 0_1$ supposed to mean?
Sorry, I guess I got a little bit carried away yesterday, hope its better now
P145: I think the last two paragraphs have a problem. Taking the intersection U\cap U_0 may lose the element $0_2$. Also, it should be enough to add a single open set to $\mathcal V$.
Suggestion below.
Informally, what's your idea for P216 (hereditarily paracompact)?
Btw, suppose $A$ and $B$ are two subspaces of a space $X$. If $A$ and $B$ are paracompact, is the union $A\cup B$ with the subspace topology from $X$ also paracompact? (My guess would be no.)
Informally, what's your idea for P216 (hereditarily paracompact)?
Btw, suppose $A$ and $B$ are two subspaces of a space $X$. If $A$ and $B$ are paracompact, is the union $A\cup B$ with the subspace topology from $X$ also paracompact? (My guess would be no.)
No even if both two subspaces are open. See https://math.stackexchange.com/questions/4754584.
Informally, what's your idea for P216 (hereditarily paracompact)?
Btw, suppose A and B are two subspaces of a space X . If A and B are paracompact, is the union A ∪ B with the subspace topology from X also paracompact? (My guess would be no.)
Same as the other proof essenitally (sps $Y \subseteq X$), given some cover of $Y$, give paracompact refinement for $Y\setminus{0_2}$ and add one neighborhood of $0_2$ intersected with some neighborhood of $0_1$ (+ union ${0_2}$) in the refinement back. (This works if both $0_1,0_2\in Y$, if one of them is missing we can use that $\mathbb{R}$ is hereditarily paraompact)