User:Lord Farin/Sandbox

This page exists for me to be able to test features I am developing. Also, incomplete proofs may appear here.

Feel free to comment.

Over time, stuff may move to User:Lord_Farin/Sandbox/Archive.

Subpages of this one may exist; they are listed at this PW special page.

Adapted converse part of Compact Subspace of Linearly Ordered Space
Let $\left({X, \preceq, \tau}\right)$ be a totally ordered space.

Let $Y \subseteq X$ be a subset of $X$ such that:

$(1): \quad Y$ is closed in $\left({X, \tau}\right)$

$(2): \quad Y$ is a complete lattice under the restricted ordering $\preceq \restriction_Y$

$(3): \quad$ For each open $U$ in the subspace topology on $Y$:
 * $\inf U \in U$ implies $U$ is lower closed in $Y$
 * $\sup U \in U$ implies $U$ is upper closed in $Y$

Here, $\inf$ denotes infimum with respect to $\preceq \restriction_Y$. (lower closed means "equals its lower closure")

Then $Y$ is compact in $\left({X, \tau}\right)$.

Proof (Sketch)
''Unlike my claim suggested, I didn't manage to make avoiding Zorn rigorous yet. Therefore, a simpler proof based on Zorn.''

Let $\mathcal U$ be an open cover of $Y$ by basis elements.

Let $\mathcal S$ be defined as:


 * $\mathcal S := \left\{{ \mathcal F : \mathcal F \subseteq \mathcal U \text{ finite}, \bigcup \mathcal F \text{ lower closed} }\right\}$

ordered by:


 * $\mathcal F_1 \preccurlyeq \mathcal F_2$ iff $\bigcup \mathcal F_1 \subseteq \bigcup \mathcal F_2$

(that this is an ordering uses essentially that the $\mathcal F$ are finite, to establish reflexivity)

Lemma: Chain Condition for $\mathcal S$
For a chain $\mathcal C$ in $\mathcal S$, let $y = \sup \left({\bigcup \bigcup \mathcal C}\right)$.

Let $U \in \mathcal U$ such that $y \in U$.

By assumption $(3)$, there must be $x \in U$ such that $x \prec y$ (or $y$ would be a smallest element of $U$, hence $\inf U = y \in U$).

Let $\mathcal F \in \mathcal C$ such that $x \in \bigcup \mathcal F$.

Define $\mathcal F' := \mathcal F \cup \{ U \}$.

Let us show that $\bigcup \mathcal F'$ is lower closed.

Let $x_1 \prec x_2$, and suppose that $x_2 \in \bigcup \mathcal F'$.

Then if $x_2 \in \bigcup \mathcal F$, $x_1$ is as well, since $\bigcup \mathcal F$ is lower closed.

Suppose now that $x_2 \in U$.

Suppose further that $x \preceq x_1$.

Then $x_1 \in U$, because $U$ is convex.

Suppose on the other hand that $x_1 \prec x$.

Then since $x \in \bigcup \mathcal F$, also $x_1 \in \bigcup \mathcal F$.

Hence $\bigcup \mathcal F'$ is lower closed, and $\mathcal F' \in \mathcal S$.

Recall that $y \in U$, hence $y \in \bigcup \mathcal F'$.

Further, by definition of $y$, we have that for all $\mathcal G \in \mathcal C$:


 * $\forall z \in \bigcup \mathcal G: z \prec y$

Since $\bigcup \mathcal F'$ is lower closed, this means:


 * $\forall z \in \bigcup \mathcal G: z \in \bigcup \mathcal F'$

and by definition of subset and $\preccurlyeq$, we conclude:


 * $\forall \mathcal G \in \mathcal C: \mathcal G \preccurlyeq \mathcal F'$

showing that $\mathcal F'$ is an upper bound for $\mathcal C$ with respect to $\preccurlyeq$.

Hence, by Zorn's Lemma, there exists a $\preccurlyeq$-maximal element $\mathcal F \in \mathcal S$.

Suppose $\sup \left({\bigcup \mathcal F}\right) \notin \bigcup \mathcal F$.

Then the construction from the lemma would extend $\mathcal F$ to an $\mathcal F' \in \mathcal S$ with $\sup \left({\bigcup \mathcal F}\right) \in \bigcup \mathcal F'$.

This contradicts maximality of $\mathcal F$.

Hence $\sup \left({\bigcup \mathcal F}\right) \in \bigcup \mathcal F$.

By condition $(3)$, $\bigcup \mathcal F$ is upper closed.

Hence $\bigcup \mathcal F$ is $Y$.

Thus $\mathcal F$ is a finite subcover of $Y$.

That is, $Y$ is compact.

Discussion

 * Do you actually use $(1)$ anywhere? --Dfeuer (talk) 17:16, 9 February 2013 (UTC)


 * Probably tacitly, or maybe it follows from $(2)$ and $(3)$. It's necessary because the ambient space is Hausdorff. --Lord_Farin (talk) 21:32, 9 February 2013 (UTC)


 * Prolonged thought suggests that $(1)$ follows from $(2)$. Furthermore, it is seen that $(3)$ is equivalent to $Y$ being connected. Since a compact set has only finitely many conn.cmpts, there is no harm in replacing it with the weaker "$Y$ has fin. many conn.cmpts". --Lord_Farin (talk) 22:06, 9 February 2013 (UTC)

New Proof of To-be-improved part of Compact Subspace of Linearly Ordered Space

 * So far I'm finding this harder to understand than my own mess. --Dfeuer (talk) 23:49, 7 February 2013 (UTC)


 * That's a pity. I think both proofs are quite accessible. --Lord_Farin (talk) 09:03, 8 February 2013 (UTC)

It's getting a bit late so I'll skip on the house style and focus on the argument - hence, this is not for verbatim publishing.

Let $Y$ be compact, and let $S \subseteq Y$.

Let us prove that $\sup S$ exists.

Since $Y$ is totally ordered, WLOG $S$ is lower closed (for the supremum won't change). to be explicated

The case that $S$ has a greatest element is trivial; suppose it hasn't.

Suppose first that $S$ has no upper bound at all.

Thus:


 * $\forall y \in Y: \exists s \in S: y \prec s$

and hence:


 * $Y = \displaystyle \bigcup \{{ {\downarrow} s: s \in S}\}$

Since $Y$ is compact, there exists a finite $S' \subseteq S$ such that:


 * $Y = \displaystyle \bigcup \{{ {\downarrow} s: s \in S'}\}$

Let $\bar s = \max S'$, which exists as $Y$ is totally ordered (invoke result).

Then for all $s \in S'$, $\bar s \notin {\downarrow} s$, hence:


 * $Y \ne \displaystyle \bigcup \{{ {\downarrow} s: s \in S'}\}$

Thus $S$ has an upper bound. Define $U = \{{y \in Y: \forall s \in S: s \prec y}\}$.

Since $S$ is lower closed, we have that $S = \displaystyle \bigcup \{{ {\downarrow} s: s \in S}\}$ is open in $Y$.

Suppose that $U$ does not have a smallest element, i.e.:


 * $\forall u \in U: \exists v \in U: v \prec u$

Then it follows that $U = \displaystyle \bigcup \{{ {\uparrow} u: u \in U}\}$ is an open cover of $U$.

Since $Y = S \cup U$ (prove), this also yields an open cover of $Y$.

Let $U' \subseteq U$ be finite such that $Y = S \cup \displaystyle \bigcup \{{ {\uparrow} u: u \in U'}\}$

Let $\bar u = \min U'$. Then $\bar u$ is not an element of the cover, a contradiction.

Thus $U$ has a smallest element, and hence $\sup S$ exists.

That $\inf S$ exists follows by duality (formalise).

Proof of Joining Arcs makes Another Arc?
I have the following idea for a proof. I'm not sure about what it needs to actually work just yet, but I feel that it's in the right direction (the correct ansatz, Germans would say).

Define $\mathcal R = \left\{{\left({t, u}\right) \in I \times I: f \left({t}\right) = g \left({u}\right)}\right\}$.

Then $\mathcal R$ is non-empty as $\left({1, 0}\right) \in \mathcal R$.

Next, we define the relation $\preceq$ on $\mathcal R$ by:


 * $\left({t, u}\right) \preceq \left({t', u'}\right)$ iff $t \le t'$ and $u' \le u$

(Include lemma proving $\preceq$ is an ordering; this is almost immediate.)
 * This is what I would like to be called the product ordering, as Kelley and at least a couple others do. --Dfeuer (talk) 19:29, 20 December 2012 (UTC)

Suppose now $\preceq$ has a minimal element $\left({t, u}\right)$.

Then $\left({t, u}\right) \ne \left({0, 1}\right)$, for $a \ne c$.

(Elaborate on why patching $f$ up to $t$ and $g$ from $u$ on yields arc.)

By continuity (I think I use Hausdorffness or something similar here), one can prove that every downward chain in $\mathcal R$ has a lower bound, so an application of Zorn's Lemma shows the minimal element exists.

When eventually putting up the argument, I'll inverse the order sign to make Zorn's lemma more intuitively applicable.

I'd be delighted if someone could avoid AC-equivalents, or point to a resource mentioning its necessity. --Lord_Farin (talk) 14:36, 15 November 2012 (UTC)


 * The words and symbols are blurring before my eyes. Headspace minimal at the moment. Can do little but routine mechanical maintenance at the moment, till this project I'm on is truly finished and (most importantly) the client has paid. --prime mover (talk) 22:05, 15 November 2012 (UTC)


 * How about this (without AoC):


 * Since $T$ is Hausdorff, [...] the diagonal set $\Delta_T$ is closed in $T \times T$.


 * Let $F: I \times I \to T \times T$ be the mapping defined by $F \left({x, y}\right) = \left({f \left({x}\right), g \left({y}\right)}\right)$.


 * By [some theorem], $F$ is continuous.


 * Not sure what that theorem might be called, but it's trivial: $f^{-1}(U) \times g^{-1}(V) = F^{-1}(U \times V)$.


 * By Continuity Defined from Closed Sets, the preimage $\mathcal R = F^{-1} \left({\Delta_T}\right)$ is closed in $I \times I$.


 * From the Heine–Borel Theorem, $\mathcal R$ is compact in $I \times I$.


 * From Continuous Image of Compact Space is Compact, $\left\{{u - v: \left({u, v}\right) \in \mathcal R}\right\}$ is compact in $\R$.


 * Note that $\mathcal R$ is non-empty, as $\left({1, 0}\right) \in \mathcal R$.


 * [...] So $\exists \left({u_0, v_0}\right) \in \mathcal R: \forall \left({u, v}\right) \in \mathcal R: u_0 - v_0 \le u - v$. (By the way, $\left({u_0, v_0}\right)$ is a $\preceq$-minimal element of $\mathcal R$.)


 * Let $\lambda = 1 + u_0 - v_0$.


 * Since $a \ne c$, [...] we have $\lambda > 0$.


 * Let $h: I \to T$ be the mapping defined by:
 * $h \left({x}\right) = \begin{cases}

f \left({\lambda x}\right) &: \lambda x \le u_0 \\ g \left({\lambda x - u_0 + v_0}\right) &: \lambda x > u_0 \end{cases}$


 * [...] Then $h$ is an arc in $T$.


 * --abcxyz (talk) 01:01, 16 November 2012 (UTC)


 * By the way, does anyone know why "Category:Proofread" shows up so big on this (specific) page? --abcxyz (talk) 01:03, 16 November 2012 (UTC)


 * Nice proof (sketch)! I really like the topological approach. It appears to be correct - it's reassuring that this argument again uses Hausdorffness, strengthening the intuition that this is the right condition on $T$. As is so often the case, when specific knowledge of the ordering is at hand, one can avoid Zorn. Here it is no different. --Lord_Farin (talk) 12:32, 16 November 2012 (UTC)


 * Thanks; as you can see, a couple of results need to be put up first.
 * It seems like the double is causing the unusually big size. I've removed one of them. --abcxyz (talk) 17:50, 17 November 2012 (UTC)

Although Steen and Seebach define an arc as ProofWiki does, some others use a narrower definition, requiring an arc to be homeomorphic to $[0,1]$. See, e.g.,. Note that attempts to point out a serious error in S&amp;S with regard to arc connectedness, but I think may misinterpret one aspect of S&amp;S's definition. A response to that post,, states that
 * every Hausdorff path-connected space is arc-connected
 * The path is compact and metrizable (Hausdorffness helps); it is also locally connected, being the quotient of [0,1], now follow the proof of the Hahn-Mazurkiewicz theorem to construct an arc in the path that connects the begin and end points.

Assuming this is right, assuming Hausdorff in this theorem renders it silly. --Dfeuer (talk) 02:50, 21 December 2012 (UTC)


 * As follows from the assumption that ref. [2] is correct, the theorem is plainly false if we drop Hausdorffness (as the teleophase topology points out (which I also covered on the talk of that page)). The whole section of the site appears to be up for review. But not by me, I don't have the sources or the patience, nor do I enjoy it. I do however feel that any reworking should take place from a thorough investigation of the literature, which is to be referenced. --Lord_Farin (talk) 15:18, 21 December 2012 (UTC)