## Closure of a subspace is in the space

Let X be a normal linear space and $$Y\subset X$$ Want to show the closure of Y is also a subspace of X

## Algebraic closure of $\mathbb{Q}$ in $\mathbb{Q}_p$

Algebraic elements in $$\mathbb{Q}_p$$ are well studied and there was a question on mathoverflow about information regarding this.

I want to know following information:

Let $$K=\overline{\mathbb{Q}}\cap \mathbb{Q}_p$$. So we have extensions

$$\mathbb{Q} \subset K \subset \mathbb{Q}_p.$$ The first one is algebraic extension (of $$\mathbb{Q}$$) and next one is transcendental extension (of $$K$$).

Q. What is known about degrees of these extensions: $$[K:\mathbb{Q}]$$ and $$[\mathbb{Q}_p:K]$$?

## Swift Memory Cycle From Closure

I am working with closures and want to confirm that I do not have any memory leaks in my code. Below is my current code:

// in ImageGalleryController.swift class ImageGalleryCollectionViewController: UICollectionViewController {   var images = [Image]()  func dropInteraction(_ interaction: UIDropInteraction, performDrop session: UIDropSession) {     // assignment of closure     var cellImage = Image() { [weak self](image) in       self?.images.append(image)       self?.collectionView.reloadData()     }     // asynchronous     session.loadObjects(ofClass: NSURL.self) { urls in       if let url = urls.first as? URL {         cellImage.url = url       }     }     // asynchronous     session.loadObjects(ofClass: UIImage.self) { images in       if let image = images.first as? UIImage {         cellImage.aspectRatio = Double(image.size.height / image.size.width)       }     }   } }  // in ImageGallery.swift struct Image {   var url: URL? = nil {     didSet {       if propertiesAreSet() { handler(self) }     }   }   var aspectRatio: Double? = nil {     didSet {       if propertiesAreSet() { handler(self) }     }   }    init(handler: @escaping (Image) -> Void) {     self.handler = handler   }    var handler: (Image) -> Void    private func propertiesAreSet() -> Bool {     let bool = (url != nil && aspectRatio != nil) ? true : false     return bool   } } 

Basically my logic is that when two separate asynchronous requests get and set two separate properties on the Image struct, then I want to append that Image struct to the images array.

Which brings me to my related questions:

• Since I am fairly new to Swift, am I properly handling memory leaks by declaring [weak self] when I create a new Image instance?
• Should I set the closure handler to nil after it executes?
• Is what I am doing ok or am I violating any Swift principles?

## Can closure and complement generate 14 distinct operations on a topological space if no subset generates more than 6 distinct sets?

$$\newcommand{\XT}{(X,\mathcal{T})}$$From Definition 1.2 in Gardner and Jackson (GJ): The $$K$$‑number $$K(\XT)$$ of a topological space $$\XT$$ is the cardinality of the Kuratowski monoid of operators on $$\XT$$ generated by closure $$b$$ and complement $$a$$ under composition. For any $$A\subset X,$$ the $$k$$‑number $$k(A)$$ of $$A$$ is the cardinality of the family of subsets generated by $$A$$ under $$\{a,b\}.$$ The $$k$$‑number of the space is $$k((X,\mathcal{T}))=\max\{k(A):A\subset X\}.$$

Question. What values does the ordered pair $$(K(\XT),k(\XT))$$ take?

Partial Answer. Lemma 2.8 in GJ states that $$k(\XT)=4$$ iff $$\XT$$ is a non-discrete door space or $$\mathcal{T}\setminus\varnothing$$ is a filter in $$2^X.$$ Each implies $$K(\XT)<14,$$ hence $$(14,4)$$ is impossible (GJ p. 25).

The case $$(14,6)$$ only gets mentioned once in GJ, on p. 28: $$\unicode{x201C}\text{We do not know of … any Kuratowski space with }k\text{-number }6.\unicode{x201D}$$ (A Kuratowski space is one that satisfies $$K(\XT)=14.)$$

It is well known that $$k(A)$$ is always even (GJ p. 15). It follows from the definitions and the identity $$bababab=bab$$ that $$2\leq k(\XT)\leq K(\XT)\leq14.$$

By Theorem 2.1 in GJ, the only possible values of $$K(\XT)$$ are 14, 10, 8, 6, 2. Clearly, $$k(\XT)=2$$ iff $$\XT$$ is discrete. Thus, besides possibly $$(14,6),$$ the pair can only be:

$$(2,2),$$
$$(6,4),$$ $$(6,6),$$
$$(8,4),$$ $$(8,6),$$ $$(8,8),$$
$$(10,4),$$ $$(10,6),$$ $$(10,8),$$ $$(10,10),$$
$$(14,8),$$ $$(14,10),$$ $$(14,12),$$ $$(14,14).$$

Each occurs in some space $$\XT$$ with $$|X|\leq7$$ (this holds for both Kuratowski monoids that satisfy $$K(\XT)=10).$$ Examples are listed below.

Theorem 2.10 in GJ states that for any $$A\subset X,$$ the family of subsets generated by $$A$$ under $$\{b,i\}$$ where $$i$$ denotes interior satisfies exactly one of 30 possible collapses of the Hasse diagram:

$$\hspace{268px}$$

The table below lists these collapses in the same order as they appear in Table 2.1 in GJ. Each entry is labeled by what I am calling the $$h$$‑number of $$A$$, or $$h(A).$$ The identity operator is denoted by $$\textsf{id}.$$

$$\begin{array}{|c|c|c|c|} \hline h(A) & h(aA) & \text{collapse} & k(A) \ \hline 1 & 1 & \varnothing & 14 \ \hline 2 & 3 & bi=ibi & 12 \ \hline 3 & 2 & ib=bib & 12 \ \hline 4 & 5 & bib=b & 12 \ \hline 5 & 4 & ibi=i & 12 \ \hline 6 & 6 & ib=ibi,\ bi=bib & 10 \ \hline 7 & 7 & ib=bib,\ bi=ibi & 10 \ \hline 8 & 9 & ib=bib,\ ibi=i & 10 \ \hline 9 & 8 & bi=ibi,\ bib=b & 10 \ \hline 10 & 11 & bi=ibi=i & 10 \ \hline 11 & 10 & ib=bib=b & 10 \ \hline 12 & 12 & bib=b,\ ibi=i & 10 \ \hline 13 & 13 & ibi=bi=ib=bib & 8 \ \hline 14 & 16 & ib=ibi=i,\ bi=bib & 8 \ \hline 15 & 17 & ib=bib,\ bi=ibi=i & 8 \ \hline 16 & 14 & ib=ibi,\ bi=bib=b & 8 \ \hline 17 & 15 & ib=bib=b,\ bi=ibi & 8 \ \hline 18 & 19 & bi=ibi=i,\ bib=b & 8 \ \hline 19 & 18 & ib=bib=b,\ ibi=i & 8 \ \hline 20 & 21 & ibi=bi=ib=bib=i & 6 \ \hline 21 & 20 & ibi=bi=ib=bib=b & 6\ \hline 22 & 22 & ib=ibi=i,\ bi=bib=b & 6 \ \hline 23 & 24 & \textsf{id}=b,\ ib=ibi=i,\ bi=bib & 6 \ \hline 24 & 23 & \textsf{id}=i,\ bi=bib=b,\ ib=ibi & 6 \ \hline 25 & 25 & ib=bib=b,\ bi=ibi=i & 4,6 \ \hline 26 & 27 & \textsf{id}=bi=bib=b,\ ib=ibi=i & 4 \ \hline 27 & 26 & \textsf{id}=ib=ibi=i,\ bi=bib=b & 4 \ \hline 28 & 29 & \textsf{id}=b,\ ibi=bi=ib=bib=i & 4 \ \hline 29 & 28 & \textsf{id}=i,\ ibi=bi=ib=bib=b & 4 \ \hline 30 & 30 & ibi=bi=ib=bib=b=i=\textsf{id} & 4 \ \hline \end{array}$$

When $$h(A)=25,$$ if $$A$$ is dense with empty interior (e.g., $$\mathbb{Q}$$ in $$\mathbb{R}$$ under the usual topology) then $$k(A)=4,$$ otherwise $$k(A)=6.$$

Conjecture. Let $$A$$ and $$B$$ be subsets of a topological space. If $$\tag1h(A)\in\{22,23,24,26,27\}\text{ and }h(B)=25,$$ then $$\tag2\min\{h(A\cup B),h(A\cap B),h(A\cup aB),h(A\cap aB)\}<20.$$

If true, the conjecture implies $$(14,6)$$ is impossible. For, suppose $$\XT$$ is a Kuratowski space with $$k$$‑number 6. Since $$K(\XT)=14,$$ there exist subsets $$A$$ and $$B$$ of $$X$$ such that $$biA\neq ibiA$$ and $$ibB\neq ibiB.$$ Since $$k(\XT)=6,$$ it follows from the table that $$(1)$$ holds. The conjecture then contradicts $$k(\XT)=6.$$

Computer experiments have verified the conjecture for all 2450 inequivalent $$\text{non-}T_0$$ spaces such that $$1\leq|X|\leq7$$ (these were recently posted here) and roughly 40,000 others such that $$8\leq|X|\leq16.$$ We ignore $$T_0$$ spaces because finite ones satisfy $$K(\XT)\leq10$$ by Theorem 3 in Herda and Metzler. We also ignore sets $$A$$ such that $$h(A)\in\{24,27\}$$ because our C program scours entire power sets and De Morgan’s laws imply that $$A$$ provides a counterexample iff $$aA$$ does.

Our list of 136 known (at the time of writing) $$4$$‑tuples $$(h(A\cup B),h(A\cap B),h(A\cup aB),h(A\cap aB))$$ satisfying $$h(A)\in\{22,23,26\}$$ and $$h(B)=25$$ can be found here.

A few weeks ago I posted a question here based on the case $$(26,25,30,12),$$ where $$h(A)=26.$$ Currently no answer has appeared.

Here is the list promised above. It is surely complete, but we lack proof. The cardinality of each space is minimal. It is assumed that $$X=\{1,\ldots,n\}$$ when $$|X|=n.$$

$$\begin{array}{|c|c|c|c|} \hline \text{pair} & |X| & \text{base for }\mathcal{T} & h\text{-numbers that occur} \ \hline (2,2) & 1 & \{\{1\}\} & 30 \ \hline (6,4) & 2 & \{\{1,2\}\} & 25,30 \ \hline (6,6) & 3 & \{\{1\},\{2,3\}\} & 25,30 \ \hline (8,4) & 2 & \{\{1\},\{1,2\}\} & 28\text{-}30 \ \hline (8,6) & 3 & \{\{1\},\{1,2\},\{1,2,3\}\} & 20,21,28\text{-}30 \ \hline (8,8) & 4 & \{\{1\},\{2\},\{1,3\},\{2,4\}\} & 13,28\text{-}30 \ \hline (10,4) & 3 & \{\{1\},\{2\},\{1,2,3\}\} & 26\text{-}30 \ \hline (10,6) & 4 & \{\{1\},\{2\},\{1,2,3\},\{1,2,4\}\} & 22,26\text{-}30 \ \hline (10,8) & 4 & \{\{1\},\{2\},\{1,3\},\{1,2,4\}\} & 14,16,23,24,26\text{-}30 \ \hline (10,10) & 5 & \{\{1\},\{2\},\{1,3\},\{2,4\},\{1,2,5\}\} & 6,14,16,23,24,26\text{-}30 \ \hline (14,8) & 4 & \{\{1\},\{2,3\},\{1,2,3,4\}\} & 18,19,26\text{-}30 \ \hline (14,10) & 5 & \{\{1\},\{2,3\},\{1,4\},\{1,2,3,5\}\} & 10,11,14,16,18,19,23,24,26\text{-}30 \ \hline (14,12) & 6 & \{\{1\},\{2\},\{3,4\},\{1,5\},\{1,2,6\}\} & 4,5,12,14\text{-}17,23\text{-}30 \ \hline (14,14) & 7 & \{\{1\},\{2\},\{3,4\},\{1,5\},\{2,6\},\{1,2,7\}\} & 1,4,5,6,12,14\text{-}17,23\text{-}30 \ \hline \end{array}$$

## stop Property Pane closure on validation fail

I am looking to find out the correct way to stop Property Pane closure or update on validation fail, I can’t see anything in the SharePoint docs and am thinking one way to do it would be to add a check on ‘onDispose’ – has anyone else implemented such functionality

## Calculate the number of distinct permutations of length n in the closure of a language

I am studying a distance CS course, but there is no tutor available, so I would appreciate your help…

Consider the language S = {a aa ab} How many distinct words of length n will appear in S*?

Is there a way to do this without having to “enumerate” S* until the answer is found?

## How to find the transitive closure of this set?

Find the transitive closure of {[a2,a2],[a5,a5],[a9,a7],[a9,a2],[a4,a5]}

over the set {a2,a4,a5,a7,a9}.

Could someone please explain step by step on how to find the solution?

Thanks

## closure of. a point

I have this topology on $$\mathbb{R}^2$$

$$\tau=\{\emptyset\}\cup \{\Omega_r, r\geq0\}$$ where $$\Omega_r=\{(x,y)\in \mathbb{R}^2, (x-2)^2+(y+2)^2\geq r^2\}$$

How to find the closure of the center $$cl(\{(2,-2)\})$$?

The only open which intersect the center is $$\Omega_0=\mathbb{R}^2$$

how to continue ?

## Orthogonal projection onto the closure of a subspace spanned by a set of orthonormal vectors

Suppose $$H$$ is a Hilbert space and $$\{e_1,\cdots\}$$ is a set of infinitely many orthonormal vectors. Then let $$M$$ be the closure of the subspace spanned by these orthonormal vectors.

I know that for $$f\in H$$, the orthogonal projection of $$f$$ onto $$M$$ is given by $$\sum_i \langle f,e_i \rangle e_i$$ However, How can I justify that the infinite sum converges?

My way is to extend these orthonormal vectors to an orthonormal basis. So we have $$\{e_i\} \cup \{h_j\}$$ as an orthonormal basis. Then $$f=\sum_i \langle f,e_i \rangle e_i + \sum_j \langle f,h_j \rangle h_j$$. So that the concerned infinite sum must converge. However, extending these vectors to an orthonormal basis will use Zorn’s Lemma. Is there any other argument to justify the convergence of $$\sum_i \langle f,e_i \rangle e_i$$?

## calculation of Transitive Closure

This question is not from homework, but rather as preparation for the test:

The calculation of place-in is a calculation in which the algorithm does not need space beyond the output size (beyond beyond a fixed space complexity). Show how, given a graph (E, V = G), the transitive closure can be calculated in-place in the calculation of G * = (V, E *)

I tried to represent the graph using a adjacency matrix. what is the next step?