show long circle is T5 but not perfectly normal

[StevenClontz]

Closes #754.

Need to justify that perfectly normal is a hereditary property first.

[prabau]

Great idea about adding hereditary in a meta-properties section for T6
(similar to what was done in https://topology.pi-base.org/properties/P000109 maybe, or whatever common format you think would be good).

Lots of references available, Engelking, mathse, etc.

[StevenClontz]

e.g. https://math.stackexchange.com/questions/1840614/perfectly-normal-is-hereditary

[Moniker1998]

Given the following equivalent definitions, that perfectly normal property is hereditary would be the easiest to show from "every closed set is a zero-set".
For if $Y\subseteq X$ and $A\subseteq Y$ is closed, then $\overline{A}$ is a zero-set of $X$ so $A = \overline{A}\cap Y$ is a zero-set of $Y$.

The argument is so short and easy, that it might be good to add it instead of a reference, even.

[Moniker1998]

Or just reference a theorem https://topology.pi-base.org/theorems/T000156

[prabau]

I agree with @Moniker1998 that to show that P15 (perfectly normal) is hereditary, it's nearly immediate if we use the characterization of P15 as "closed sets are zero sets". That is in fact the proof in Engelking 2.1.6. So the easiest would be to quote that theorem. That way, we don't have to spend a lot of real estate on the page to validate the meta-property.

I don't like the https://math.stackexchange.com/questions/1840614/perfectly-normal-is-hereditary too much as a reference, because the poster is sloppy. His definition of perfectly normal is not even correct.

[prabau]

The arguments for the S196 traits are insufficient I think.

For P15, we should mention that perfectly normal is a hereditary property (without justification).

For P8 (T5), we also need to show the whole space is T4.
Also, can you explain to me why every proper subspace is a LOTS?

[prabau]

To show that the long circle is T5, we need:

• show $X$ is T4
• show every proper subspace is T4, it's enough to show every subspace obtained by removing a single point is T5. That's because it's a LOTS, as you mentioned. To see this, the argument in https://topology.pi-base.org/spaces/S000196/properties/P000204 gives the idea.

[yhx-12243]

I think we can show instead Long circle (S196) is not countably tight (P81), and thus is not perfectly normal automatically.

https://github.com/pi-base/data/pull/990/files#diff-f42ba7d033486c9e0008cb8cb76cb858df0ca916f52c3d1d2609d5ef097b97ce

(You can review and cherry-pick that file from #990).

[yhx-12243]

[yhx-12243]

To show that the long circle is T5, we need:

• show $X$ is T4
• show every proper subspace is T4, it's enough to show every subspace obtained by removing a single point is T5. That's because it's a LOTS, as you mentioned. To see this, the argument in https://topology.pi-base.org/spaces/S000196/properties/P000204 gives the idea.

For $X$ is $T_4$, we take any two disjoint closed $A, B$. By connectivity there exists $p \in X \setminus (A \cup B)$. Then we use the normality of $X \setminus \{p\}$ to get $A \subseteq U$ and $B \subseteq V$. Then $U \setminus \{p\}$ and $V \setminus \{p\}$ is what we want.

[prabau]

@yhx-12243 FYI, if you want to show braces in mathematics formula here, one needs to use double backslashes:
$\\{x\\}$ giving $\{x\}$, and not $\{x\}$.

[StevenClontz]

Thanks all for the feedback. Working on this now.

[prabau]

S196 P8 (T5 property): This is not enough.

As explained before, it's not because every proper subspace is T5 that the whole space will be. Also need to show that the whole space is T4. (Right now, T4 follows from T5, which is circular reasoning.)

Also, the justification that every proper subspace is a GO-space is unsupported. Need to show why, or refer to some existing mathse post.

[pzjp]

It is elementary that after removing arbitrary point we obtain LOTS. And the space cannot be covered by two nonempty closed sets (connectedness), hence they are contained in a LOTS which is T4.
(Also this is a loop-ordered space mentioned in this paper )

[prabau]

It is elementary that after removing arbitrary point we obtain LOTS. And the space cannot be covered by two nonempty closed sets (connectedness), hence they are contained in a LOTS which is T4. (Also this is a loop-ordered space mentioned in this paper )

I am not saying it's difficult. But something more should be mentioned for P8.

Interesting paper by the way. I need to find the time to look at it in more detail.

[StevenClontz]

Discussion of complete normality is now factored out to https://math.stackexchange.com/questions/5017774

[prabau]

Thanks for adding the mathse post.