## messages – How to completely reset iMessage on MacOS X Catalina 10.15.4?

I have an iMac with two user accounts. After updating to MacOS Catalina 10.15.4, one of the two users can no longer use iMessage. iMessage simply does not start. Startup results

• iMessage immediately becomes "unanswered" without ever displaying a window.
• It can be "quit immediately"
• User can't sign out until iMessage has "quit immediately"
• Works great for the other user on this iMac
• Works great on all other devices (iOS and other MacOS devices of the affected user)
• iMessage on iCloud was disabled for this user during update 10.15.4
• When iMessage is not started (and crashed), the badge on the icon is correctly updated with the number of unread messages!

My idea was to completely reset iMessage to factory condition, but I don't know how to do it. What I have tried

• Logout and reconnection to iCloud
• Restoring this user from a Time Machine backup
• Remove preferences and IdentityServices as in this question and this thread (there is no McAffee used here).

The behavior is all the same. The only thing I haven't gone through is to let the user start from scratch with a new user account. It would be quite painful.

## html – Can you put all the text in a static tag for SEO but completely rewrite the page with JavaScript?

Why is there initial text in the tags? If it is for SEO reasons, then it is hidden text and definitely a bad idea for SEO.

To answer your JS question, it is good to dynamically fill the text of your page with javascript. (Google spiders with Chrome and runs JS).

The problem is really the preexisting text and why it is there. Is it different from the text you add with JS? Is it rich in keywords?

## regions – Why can't the output of the `DiscretizeRegion` function be displayed completely?

``````mr = DiscretizeRegion(Region(Rectangle({0, 0}, {Pi, Pi})),
AccuracyGoal -> 4)

r6 = TransformedRegion(mr,
Function({t,
p}, {2 Sin(p)^2 Sin(t)^2 + 5 Cos(p)^2 Sin(t)^2 -
Cos(t)^2, (4 Sin(p)^2 Sin(t)^2 +
25 Cos(p)^2 Sin(t)^2 - (2 Sin(p)^2 Sin(t)^2 +
5 Cos(p)^2 Sin(t)^2 - Cos(t)^2)^2 + Cos(t)^2)^0.5`}));
Region(r6, PlotRange -> {{-1, 5}, {0, 3}})

DiscretizeRegion(r6, PlotRange -> {{-1, 5}, {0, 3}})
``````

Why the release of `DiscretizeRegion` is the function of the above code displayed completely?

Also, I don't know how to change the display size of the downloaded image.

## Completely Free 2019 Patriots Company Profile: Danny Shelton Could Provide Low Cost Defensive Line Details

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## html – My page does not completely fill the window on mobile. Why?

``````* {margin: 0 auto; padding: 0 auto;}

/*
Por a largura se repetir
utilizei a classe default-structure-formatting
para a definir e evitar repetições.
Os atributos específicos de altura e cor estão em cada respectiva tag.
*/

.default-structure-formatting
{
width: 100%;
}

{
background-color: #000;
height: 120px;
}

nav
{
background-color: #09f;
height: 30px;
}

section
{
background-color: #891e2b;
height: 600px;
}

footer
{
background-color: #000;
height: 30px;
}

/* Formatação de Texto*/
.default-text-formatting {text-align: center; color: white; }

.default-content-formatting {width: 994px;color: white; padding: 100px auto;}

article p {padding: 220px 0; text-align: justify;}``````
``````

LOGO

HOME - PORTFOLIO - SOBRE - CONTATO

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``````

## [ Politics ] Open question: Why did the Republicans refuse to vote for Sarah Palin slightly delayed when they voted for Donald Trump completely delayed? Are Republicans Sexist?

[Politics] Open question: Why did the Republicans refuse to vote for the slightly delayed Sarah Palin when they voted for the completely delayed Donald Trump? Are Republicans Sexist?

## Magisk how to completely hide Magisk Manager and Unhide Magisk

I configure the mobile for my mom, how to completely hide Magisk Manager. I don't want to see this app. All I need to do is reveal it from the ssh or adb session.

## general topology – \$ CA \$ implicit in Munkres' proof? (Completely regular spaces)

Recently, I read the topology of Munkres. In the chapter $$4$$, he presents some proofs of the metrization theorem of Urysohn then, at the end of the section, generalizes one of them to an “ incorporation theorem '' # 39; for completely regular spaces (Munkres definition of completely regular includes the $$T_0$$ axiom):

Theorem 34.2 (nesting theorem). Let $$X$$ to be a space in which the sets of a point are closed. Assume that $${f _ { alpha} } _ { alpha in J}$$ is an indexed family of continuous functions $$f_a: X rightarrow mathbb R$$ fulfilling the requirement that for each point $$x_0$$ of $$X$$ and each neighborhood $$U$$ of $$x_0$$, there is an index $$alpha$$ such as $$f _ { alpha}$$ is positive to $$x_0$$ and disappears outside $$U$$. Then the function $$F: X rightarrow mathbb R ^ J$$ Defined by $$F (x) = left (f _ { alpha} (x) right) _ { alpha in J}$$
is a recess of $$X$$ in $$mathbb R ^ J$$. Yes $$f _ { alpha}$$ maps $$X$$ in $$(0.1)$$ for each $$alpha$$, then $$F$$ impregnated $$X$$ in $$(0,1) ^ J$$.

My question is this: is the proof of this theorem (the way Munkres describes it above) implicitly based on the axiom of choice or on a weaker version? If the answer is "a weaker version", what is this version? Is the theorem itself equivalent to a well-known weaker version of $$AC$$?

preparation:

The essential elements of the evidence – that $$F$$ is injective, continuous and an open map, follow easily by simply plugging in $$J$$ instead of the index set in the first proof of Urysohn's metrization theorem presented by Munkres.

The thing that has bothered a recent rereading of this evidence is the construction of the & # 39;$$F$$& # 39; used in the statement of the nesting theorem itself.

It seems that, implicitly, in the construction of the function, & # 39;$$F$$& # 39; we used the truth of a statement that would look something like:

Yes $$Y$$ is a set and $${Z _ { beta} } _ { beta in J}$$ is an indexed family of s.t. sets for each $$beta in J$$, $$exists$$ some $$g _ { beta}: Y rightarrow Z _ { beta}$$, then there is a function: $$G: Y rightarrow prod _ { beta in J} Z _ { beta}$$ Defined by $$G = left (g _ { beta} right) _ { beta in J}$$

But this statement looks, at least to me, a version of the Axiom of Choice.

For, with the truth of the Axiom of Choice, given these $$Y$$, $${Z _ { beta} } _ { beta in J}$$, and $${g _ { beta} } _ { beta in J}$$, we can order the index set well, $$J$$, then define recursively $$pi _ { beta} (G) = g _ { beta}$$ (or $$pi _ { beta}$$ denote it $$beta$$projection map).

On the other hand, if the above statement was true, given any indexed family of non-empty sets, $${X _ { alpha} } _ { alpha in J}$$, we have this, for each $$alpha$$, $$exists$$ $$f _ { alpha}: {0 } rightarrow X _ { alpha}$$ (since each $$X _ { alpha}$$ is not empty – I think this collection of functions should be well defined). So by the truth of the statement, there is a map, $$F: {0 } rightarrow prod _ { alpha in J} X _ { alpha}$$. But then $$prod _ { alpha in J} X _ { alpha}$$ must contain $$F (0)$$doing $$prod _ { alpha in J} X _ { alpha}$$ not empty.

So the statement seems to imply the Axiom of Choice too, making it look as if they were equivalent. I must say, however, that I am not sure that the above implications are true. My set theory is rusty and I have not found and I do not remember ever seeing the Axiom of Choice spelled out this way, so I think there may be a fault in my reasoning.

So my question is – well, first, if there is a flaw in my reasoning establishing an equivalence between the statement and $$AC$$, please leave a response explaining this – but beyond that, may the statement be equivalent to something weaker? If not, is there a way around $$AC$$ in the proof of Imbedding's theorem? I ask mainly because I don't really know $$AC$$ appearing in topology as versions of Tychonoff's theorem, the connection of which seems more intuitive.

## applications – Is there a way to completely prevent an application from using audio?

Whenever I open an app (in this particular case, Pokemon Go), my Bluetooth headphones are automatically exchanged on my phone, even if I actively use them to listen to music from another source. I never need Pokemon Go audio. Already. But I don't want to completely disable bluetooth because it runs my watch and another device as well. Currently, my solution is to disconnect my headphones from my phone every time I sit at my desk and re-pair them when I walk away, which is a huge pain. Is there a way to completely block this app for using multimedia audio, or maybe specifically Bluetooth audio? The Bose connect app has no way to set a priority list or prevent certain apps from using it.

I have already tried to revoke the permission to take audio focus (https://www.xda-developers.com/pokemon-go-audio-focus/), but I believe this refers to take it from another application and may be out of date for my device running Android 9 (Razer Phone 2).

Anyone have any ideas or suggestions?

## Stay Away From Hetzner – Outdated Systems, Completely Faulty Support Workflow

I still have 3 server contracts on Hetzner. Mainly because I didn't have time to move them.
Recently had problems with 2 of these servers. Regular broadcasts. One is SPAM (occurs all the time with clients who don't follow the rules) and a port scan on another (from time to time, this also happens if the client is using old software that , in a way, allows code injection or something like that). Who has never had this kind of problem on clients' servers? What did Hetzner do in both situations? They simply block servers by removing them from the network. You cannot access it in any way. You are trying to request access in order to verify and resolve the issue using the support interface but you are not allowed as they refuse the support form with the message "the server is blocked. Correct the problem in order to continue". You try the instruction on an automatic email, they fail (the links do not exist). You send an email to support@hetzner.de, the stupid guy on the other side, copy and paste the answer "can't do anything until you solve the problem". The server is not accessible. How the hell do these stupid people want me to fix the problem?!? So they block your server, they always ask to repair it by telepathy (I suppose ?!) But they continue to load you. Excellent service (for them only).

Stay away from these boys. Look at OVH, where I have all my other servers and move these 3. They have a modern, efficient and autonomous way of dealing with these common problems. They block dangerous traffic by type and / or port. Never move the server away from the network and do not block all traffic. And even if you have an access problem (let's say the server is inaccessible due to a driver problem or a faulty drive or something like that) you can still restart the server in "rescue mode", you can therefore immediately access it remotely using an SO flash, so that you can mount your disks, find the problem, solve it and put it back online. It all depends 100% on you and stand-alone systems only. No support required. Solution available 24 hours a day. You don't have to deal with foolish people.
How can Hetzner still be on the market with such a support staff and an obsolete system and methods? This does not correspond to the German rigor that we are used to in other areas of activity … I will stay away from these guys.