S149 and related embed into a W-group and are proximal

[Moniker1998]

No description provided.

[prabau]

Sorry, but that does not seem that easy. I thought the author of the PR was the one to choose that label?

[prabau]

Slightly confused about the notation: $1_\alpha$ versus $1_{\{\alpha\}}$ ?
Can you clarify some more?

[yhx-12243]

α = [0, α). Maybe a little bit abuse.

[prabau]

You mean the point $\alpha=\left<\alpha,0\right>$ right?

[prabau]

also why one index is as a set $\{\alpha\}$ and the other not?

[prabau]

Also, the embedding for the closed long ray does not work because the right end point needs $\alpha=\omega_1$.

In all the above, $\alpha$ is an ordinal in $\omega_1$, right?

In the description of the Long ray, $X_0=\omega_1 \times [0,1)$. So I imagine you a using a typical element $(\alpha,t)$ with $\alpha\in\omega_1$ and $0\le t<1$.
And $1_\alpha$ would be the element $1\in\mathbb R$ in the $\alpha$-th coordinate?

[Moniker1998]

Slightly confused about the notation: $1_\alpha$ versus $1_{\{\alpha\}}$ ? Can you clarify some more?

It's the indicator function

[Moniker1998]

α = [0, α). Maybe a little bit abuse.

Those are the same. Yes

[Moniker1998]

You mean the point α = ⟨ α , 0 ⟩ right?

No

[Moniker1998]

also why one index is as a set { α } and the other not?

Because those are different sets

[Moniker1998]

Also, the embedding for the closed long ray does not work because the right end point needs α = ω 1 .

That was abuse of notation. You should have interpreted $f(\omega_1, 0)$ as $1_{\omega_1}$.

In all the above, α is an ordinal in ω 1 , right?

$\alpha$ is an ordinal in $\omega_1+1$

In the description of the Long ray, X 0 = ω 1 × [ 0 , 1 ) . So I imagine you a using a typical element ( α , t ) with α ∈ ω 1 and 0 ≤ t < 1 . And 1 α would be the element 1 ∈ R in the α -th coordinate?

In the description of long ray $X = \omega_1\times [0, 1)$, not $X_0$. No, $(\alpha, t)\in X_0 = \omega_1\times [0, 1)\cup \{(\omega_1, 0)\}$. And no, here $1_\alpha$ would be the element $1_\alpha\in \prod_{\beta &lt; \omega_1}\mathbb{R}$ with $1$ in the coordinates $\beta &lt; \alpha$.

[Moniker1998]

@prabau see comments above.

[Moniker1998]

Actually it'd be better to include this result for S149 and then for all subspaces of it we would just include it also.

[Moniker1998]

There is nothing ambiguous here.

[prabau]

right, you explained it to me here. But it's too tricky for the reader without explanation on the web page. The final web page needs to be much clearer. If you want, I can expand what you have to make things clearer. (There are other things that can be made clearer too.)

[Moniker1998]

@prabau it's explained there. The notation $1_Y$ for $Y\subseteq \omega_1$ denotes the indicator function.
So "without explanation" is just not true.

[prabau]

Also, why is this map closed? It's because $X_0$ is compact, mapping continuously into a Hausdorff space. Something more should be mentioned, at least that $X_0$ is compact.

[prabau]

@prabau it's explained there. The notation 1 Y for Y ⊆ ω 1 denotes the indicator function. So "without explanation" is just not true.

It's too subtle. I'll modify things and then you can discuss later.

[prabau]

I have added clarifications for P149. What do you think?

For P76 (proximal), we need the image of the embedding into the Sigma-product to be closed in it. The reason is that $X$ is dense in $X_0$ (and $f$ is a closed map), right? Or is there a different reason?

@felixpernegger fyi

[Moniker1998]

@prabau it doesn't matter that $X$ is dense in $X_0$

[Moniker1998]

I have added clarifications for P149. What do you think?

I don't like it a lot

[prabau]

I have added clarifications for P149. What do you think?

I don't like it a lot

@felixpernegger Can you take a look? Do you think it's fine, or too verbose? (see earlier comments)

[Moniker1998]

Here, for $Y\subseteq\omega_1$ the function $1_Y:\omega_1\to\mathbb R$,
which is an element of $\prod_{\beta\in\omega_1}\mathbb{R}$, is the characteristic function of $Y$.
In particular, for an ordinal $\alpha={\gamma:\gamma&lt;\alpha}$,
$1_{{\alpha}}(\beta)=1$ iff $\beta=\alpha$ and $1_\alpha(\beta)=1$ iff $\beta&lt;\alpha$.

This is the first part I don't like. It feels almost insulting on the readers intelligence. It explains things that don't need to be explained. This is the feelings I'm getting from it.

The function $f$ is continuous and injective. Since $X_0$ is compact, $f$ is also closed and hence an embedding.
The restriction of $f$ to $X$ is an embedding into the $\Sigma$-product $W=\Sigma_{\alpha &lt; \omega_1}\mathbb{R}$
and $f(X)$ is closed in $W$.

And this is the second part I don't like. First the $W$ here is completely unnecessary. And second, $f(X)$ is closed in $W$. The previous formulation was way better.

[prabau]

First the W here is completely unnecessary. And second, f ( X ) is closed in W .

I don't necessarily disagree. But a basic question first, to convince myself.

Given a closed continuous map $f:X\to Y$ between topological spaces and a subspace $A\subseteq X$, not assumed to be closed in $X$. In general, $f(A)$ will not be closed in $Y$. Assume also there is a subspace $Z\subseteq Y$ such that $f(A)\subseteq Z$.
What conditions guarantee that $f(A)$ will be closed in $Z$?

(I thought I convinced myself when $A$ was dense in $X$. But please confirm and check if it's not needed.)

[Moniker1998]

$f(A) = f(X)\cap Z$

[prabau]

Are you sure about that? Why is the right side contained in the left side?

It is false in general. In the special case of $Z=Y$, it's clear to see that $f(A) is in general not closed in $Y$.
So some extra condition is necessary. One cannot say that just because $f$ is a closed map, so will be the restriction of $f$ to $A$ (when considered as mapping into some restricted codomain as well).

[Moniker1998]

@prabau Am I sure about what? This is exactly the setting here. Here $f(A) = f(X)\cap Z$ which is obviously closed in $Z$ by definition of subspace topology

[prabau]

I know it is the case in this PR.

I was asking if it true in more general circumstances, and the answer is no. And what is it that makes it true in the case of this PR?
And what that is would be worthwhile to indicate. For example, the fact that $X$ is dense in $X_0$ if that is what implies it.

[prabau]

Or is it just by inspection, because $X_0\setminus X$ consists of just two points, and the two points map outside of the $\Sigma$-product ?

[Moniker1998]

Or is it just by inspection, because $X_0\setminus X$ consists of just two points, and the two points map outside of the $\Sigma$-product ?

Yes you just look at the map

[prabau]

Yes you just look at the map

Ok. That's pretty simple then. I mistakenly inferred that it was following from a general theorem about closed maps, which I did not find true.

[Moniker1998]

@prabau I like the current version

[prabau]

Apart from that, it all looks good to me. Leaving the final approval to @felixpernegger.

[felixpernegger]

Apart from that, it all looks good to me. Leaving the final approval to @felixpernegger.

I have lots of things to do tonighr, and dont have time for pibase. I will looks at this tomorrow hopefully

[felixpernegger]

I think this is fine. It is obviously a bit long, but nothing too critical since the argument still is simple. It seems very readable and not confusing to me.