Give some spaces the Toronto trait

[felixpernegger]

No description provided.

[prabau]

and same typo in the other files

[prabau]

another simple theorem that you could add, maybe in a different PR
ordinal + Toronto => discrete.

That would allow to derive that many ordinal spaces are not Toronto, specifically all ordinal $\alpha$ with $\alpha>\omega$. In particular, $\alpha=\omega+\omega+1$ (S34), for which toronto is unknown up to now.

Right now, it is known for some ranges of ordinals:

• $\alpha\le\omega$: true because discrete (using T817)
• $\omega<\alpha\le\omega+\omega$: false because almost discrete (using T844)
• $\omega_1\le\alpha<\omega_2$: false using T843

[prabau]

Not sure why you mention ordinal recursion in the justifications. That seems way overblown. And also, maybe not enough in itself.

S199 for example: For myself, what is the real reason things work here?
Giving an infinite subspace $Y\subseteq X$, the smallest nbhd of a point $y_0\in Y$ in the subspace topology will be the set $Y\cap\{x\in X:x\le y_0\}=\{y\in Y:y\le y_0\}$ (side note: Alexandrov subspace of Alexandrov is Alexandrov), which is the left ray topology on $Y$ (wrt to the induced order on $Y$). So $Y$ is homeomorphic to $X$.
(I am NOT saying we need to put all these details.)

In that a general consequence for any order preserving bijection (or rather injection?) for any linearly ordered set as you seems to indicate?
(it's certainly true in the case of infinite subsets of $\omega$)

[felixpernegger]

another simple theorem that you could add, maybe in a different PR ordinal + Toronto => discrete.

That would allow to derive that many ordinal spaces are not Toronto, specifically all ordinal α with α > ω . In particular, α = ω + ω + 1 (S34), for which toronto is unknown up to now.

Right now, it is known for some ranges of ordinals:

* α
    ≤
    ω
  : true because discrete (using T817)

* ω
    <
    α
    ≤
    ω
    +
    ω
  : false because almost discrete (using T844)

* ω
      1
    
    ≤
    α
    <
    
      ω
      2
    
  : false using T843

I wanted to avoid theorems spaces which are T2 since this may follows from more general, but sure why not,

Not sure why you mention ordinal recursion in the justifications. That seems way overblown. And also, maybe not enough in itself.

S199 for example: For myself, what is the real reason things work here? Giving an infinite subspace Y ⊆ X , the smallest nbhd of a point y 0 ∈ Y in the subspace topology will be the set Y ∩ { x ∈ X : x ≤ y 0 } = { y ∈ Y : y ≤ y 0 } (side note: Alexandrov subspace of Alexandrov is Alexandrov), which is the left ray topology on Y (wrt to the induced order on Y ). So Y is homeomorphic to X . (I am NOT saying we need to put all these details.)

In that a general consequence for any order preserving bijection (or rather injection?) for any linearly ordered set as you seems to indicate? (it's certainly true in the case of infinite subsets of ω )

The topology of left rays on some ordinal $\omega$ (and thus on a subspace) are the rays $[0,\alpha)$ for some $\alpha < \omega$. Any order preserving bijection between the space and some $\alpha \leq \omega + 1$. Any order preserving bijection map between a space and subspace sends these open rays to open rays and vice versa, and is thus a homeomorphism.

Such a bijection exists, if $X$ is a cardinal and $Y$ a subspace of same cardinality. Namely construct it by:
$f(x) = \inf\{y \in Y | y \notin f((0,x))\}$ (this exists by well orderedness. Technially we would need cutoff condition, but this is redundant by the next argument).

This $f$ clearly satie
sfies $f(x)\leq x$.
If $f$ wasnt injective, there exists a ordinal $p \in X$, with $|p|=|Y|=|X|$, contradicting $X$ bein a cardinal.
Similarly, if $f$ wasnt surjective, $f$ is (can be extended to) an injective map onto some $p \in X$ and thus again $|p|\geq |Y|=|X|$, contradicting $X$ being a cardinal.

[prabau]

I wanted to avoid theorems spaces which are T2 since this may follows from more general, but sure why not,

What results for T2 spaces are you thinking about? The Brian paper does not seem to mention any ZFC result for general Hausdorff spaces, only some for "HATS" (of size $\aleph_1$).

[felixpernegger]

I wanted to avoid theorems spaces which are T2 since this may follows from more general, but sure why not,

What results for T2 spaces are you thinking about? The Brian paper does not seem to mention any ZFC result for general Hausdorff spaces, only some for "HATS" (of size ℵ 1 ).

In #1549 (comment) @yhx-12243 claims the main result (on which most of the rest is built on) holds for any T2 Toronto space, but I dont know why (since the proof in the paper kind of depends on it).

[prabau]

(minor thing for next to last paragraph in #1621 (comment):
to display latex braces in github comments, need to duplicate the backslashes: \\{...\\} instead of \{...\} )

[prabau]

In #1549 (comment) @yhx-12243 claims the main result (on which most of the rest is built on) holds for any T2 Toronto space, but I dont know why (since the proof in the paper kind of depends on it).

I see. @yhx-12243 says that there are some general results for some general classes of t2 spaces. For example:

• toronto + T2 + scattered => discrete
• toronto + T2 + sequentially discrete => discrete

That would indeed be very good to have, and obviate the need for my suggested result for ordinals.

[felixpernegger]

On another note, our current implementation, T219, of Theorem 6.1 in W.R. Brian paper is quite insufficient (see this). Maybe we should add some more corollaries from Theorem 6.1 (i.e. T0, which can be done directly also)

[prabau]

Going back to the topic of left ray topologies, starting with an arbitrary ordered set $(X,\le)$. A subset $Y\subseteq X$ is equipped with the order induced from the order of $X$. $X$ has a left ray topology corresponding to its order, and so does $Y$ for its corresponding order.

Now suppose $X$ is a cardinal (i.e., an initial ordinal), viewed as a set of ordinals as usual. Then, if $|Y|=|X|$, the two ordered sets $(X,\le)$ and $(Y,\le)$ are isomorphic (by a unique bijection, easily constructed using well-orderedness and what you mentioned). And of course such an order preserving bijection also preserves the corresponding left ray topologies. But this is NOT what needs to be proved.

Instead, we need to show that the left ray topology on $(Y,\le)$ coincides with the subspace topology induced from the topology of $X$.

It's not too difficult to do. I don't feel like writing down the whole thing, but for one part of it one can use the fact that $Y$ must be cofinal in $X$ and the well-order property to show that if $x\in X\setminus Y$, there is a smallest $y\in Y$ such that $x<y$ and then $(\leftarrow,x)_X\cap Y = (\leftarrow,y)_Y$.

But the justification in the PR (By ordinal recursion, we can construct a (unique) order preserving bijection $f:Y \to X$ which then must be a homeomorphism.) is somewhat misleading.

[felixpernegger]

Going back to the topic of left ray topologies, starting with an arbitrary ordered set ( X , ≤ ) . A subset Y ⊆ X is equipped with the order induced from the order of X . X has a left ray topology corresponding to its order, and so does Y for its corresponding order.

Now suppose X is a cardinal (i.e., an initial ordinal), viewed as a set of ordinals as usual. Then, if | Y | = | X | , the two ordered sets ( X , l e ) and ( Y , l e ) are isomorphic (by a unique bijection, easily constructed using well-orderedness and what you mentioned). And of course such an order preserving bijection also preserves the corresponding left ray topologies. But this is NOT what needs to be proved.

Instead, we need to show that the left ray topology on ( Y , ≤ ) coincides with the subspace topology induced from the topology of X .

It's not too difficult to do. I don't feel like writing down the whole thing, but for one part of it one can use the fact that Y must be cofinal in X and the well-order property to show that if x ∈ X ∖ Y , there is a smallest y ∈ Y such that x < y and then ( ← , x ) X ∩ Y = ( ← , y ) Y .

But the justification in the PR (By ordinal recursion, we can construct a (unique) order preserving bijection $f:Y \to X$ which then must be a homeomorphism.) is somewhat misleading.

Yes, I thought its somewhat obvious the subspace topology is the left ray topology, so I didnt mention it explicitly. I dont think we have to, or at least not prove it. hm

[prabau]

For S199 and S200 (left and right ray topologies on $\omega$), it's king of obvious though. So feel free to simplify things slightly, and no need to don't mention "ordinal recursion" for this case.

And for the other space $\omega_1$, maybe an easier reason why things work is that for ordinal spaces, the left ray topology = "left open ray" topology coincides with the "left closed ray" topology (because $ (\leftarrow, \alpha]$ = $(\leftarrow,\alpha+1)$ ). From that, it's even easier to see that the two involved topologies for $Y$ coincide.

Not sure about "right ray" topologies (open ray or closed ray would be different).

[prabau]

Yes, I thought its somewhat obvious the subspace topology is the left ray topology, so I didnt mention it explicitly. I dont think we have to, or at least not prove it. hm

We don't need to give details of the proof, but the text needs to be changed to not give misleading ideas.

[prabau]

I have created #1622. In particular, $X=$ the right "open ray" topology on $\omega_1$ has the property that every subset $Y$ with the same cardinality as $X$ is order isomorphic to $X$. And yet, $X$ is not Toronto. So that will be good to illustrate things.

Apart from that, let me try to make a few suggestions for this PR.

[felixpernegger]

thanks

[prabau]

Suggested change

	Let $Y\subseteq X$ with $|Y|=|X|$. By ordinal recursion, we can construct a (unique) order preserving bijection $f:Y \to X$ which then must be a homeomorphism.
	Let $Y\subseteq X$ with $|Y|=|X|$.
	There is a unique order isomorphism $f:Y\to X$.
	Since $X$ is Alexandrov, so is $Y$ with its subspace topology $\tau_Y$.
	The smallest $\tau_Y$-neighborhood of a point $y\in Y$ is
	$$(\leftarrow,y]_X\cap Y = (\leftarrow,y]_Y,$$
	which coincides with the smallest neighborhood of $y$ in the left ray topology of $Y$.
	This shows that $f$ is a homeomorphism.

[felixpernegger]

What if we just say that the subspace of topology of $Y$ is the same as its left ray topology without proof?

[prabau]

Yeah, I think that should be fine for S199 and S200.

[prabau]

Suggested change

	Let $Y\subseteq X$ with $|Y|=|X|$. By ordinal recursion, we can construct a (unique) order preserving bijection $f:Y \to X$ which then must be a homeomorphism.
	Let $Y\subseteq X$ with $|Y|=|X|$.
	There is a unique order isomorphism $f:Y\to X$.
	Since $X$ is Alexandrov, so is $Y$ with its subspace topology $\tau_Y$.
	The smallest $\tau_Y$-neighborhood of a point $y\in Y$ is
	$$[y,\to)_X\cap Y = [y,\to)_Y,$$
	which coincides with the smallest neighborhood of $y$ in the left ray topology of $Y$.
	This shows that $f$ is a homeomorphism.

[prabau]

Suggested change

	Let $Y\subseteq X$ with $|Y|=|X|$. By ordinal recursion, we can construct a (unique) order preserving bijection $f:Y \to X$ which then must be a homeomorphism.
	Let $Y\subseteq X$ with $|Y|=|X|$.
	Using well-orderedness and the fact that $\omega_1$ is an initial ordinal,
	one can show that there is a unique order isomorphism $f:Y\to X$.
	Since $X$ is Alexandrov, so is $Y$ with its subspace topology $\tau_Y$.
	The smallest $\tau_Y$-neighborhood of a point $y\in Y$ is
	$$(\leftarrow,y]_X\cap Y = (\leftarrow,y]_Y,$$
	which coincides with the smallest neighborhood of $y$ in the left ray topology of $Y$.
	This shows that $f$ is a homeomorphism.

[yhx-12243]

What about just saying that “Similar to the proof of S199”?