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You got the right answer, but

must be real.

Note that

is on the unit circle when are real, so it has absolute value 1.Therefore,

Set

so and so to getTherefore,

and thus,

is normalized

Use mathematical induction

since

Assume

Then

so

since

shows it cannot be negative

Hence, by induction we have

for all positive integers n

Avon wrote:

The order of quantifiers is important.

betterthangauss wrote:

Therefore, for any neighborhood of , there is an integer , , such that

This is true but the value of many depend of the neighbourhood .

What you want to show is the there exists an integer such that for any neighbourhood of of

Thanks. After taking a second look at the problem I saw a really easy way to prove it. Namely:

Therefore,

Note that

is the finite union of closed sets, so it is closed (this doesn't hold in the countable case).

Therefore, since

is the smallest closed set containing , we haveThanks again for showing me my error.

Let me just post images instead, so that it's easier to read:

QUESTION

MY ANSWER

NOTE: By J_n I mean the set

**betterthangauss**- Replies: 3

Let

be a finite collection of sets in a metric space and setThen the closure of

, namely , satisfiesThe union of the closures of A_i's being a subset of the closure of B is easy to show. However, the reverse direction is tripping me up a bit. Here's how I argued it:

Assume

Then is adherent to

Then for any neighborhood of , we have

i.e., for any neighborhood

of , we haveTherefore, for any neighborhood

of , there is an integer , , such thatHence, there is an integer

, , such that.

Therefore,

Hence,

Is this argument fallacious? I ask, because I can't see where the finite part of the union comes into play. It seems like I could use this exact same argument to say the result holds in the countable case too, but there is an obvious counterexample to that:

Namely, setting

to the set of the rational numbers and each to the singleton set where (since is countable).Then the closure of each

is just the set , since each is just the singleton ., the set of real numbers, while.

Hence, the two sides are not equal.

Can someone show me where this first argument above was going bad? Thanks.

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Ricky wrote:

It's a good command to have, but with a pretty long name. Reminds me of Java. I would recommend:

SetC

Personally in commands, the more lowercase the better so I would actually make it

setc

It's all about reducing the number of keystrokes (whilst maintaining readability).

When I'm coding something in any language, I only like to use short variable names for names defined in small scopes. For something global that's going to be used a lot to express the semantics of my paper, I'd much rather be as explicit as possible in my naming. I feel like if I stop writing a document and then come back a year later to make corrections or changes, I'll understand \SetWithConstraint more quickly than I will \SetC (maybe I'd think \SetC was a command for the set of complex numbers, for instance). I use the camel-case with uppercase first letters to try to keep my defined names from conflicting with the symbols AMS defines.

math. Using \left to line up the | doesn't look right when the right side is small and the left side big

i.e., here's the code for the command:

`\newcommand{\SetWithContraint}[2]{\left\{\,#1\ \left|\ #2\,\right.\right\}}`

And if I use it like this:

```
\[
\Defined
{B_N}
{\SetWithContraint
{
\SetMember
{
\FiniteSum{a_i z^{n-i}}{i}{0}{n}
}
{
\Polynomials{\ComplexNumbers}{z}
}
}
{
FOO!
}
}
\]
```

I get

where \SetMember, \Defined, etc. are commands I made up to try to decouple the logic of the markup from the characters used to display it.

The following recursive algorithm will do it without any table lookup, and could be done by hand without having to do anything but multiply, subtract, and divide

It should be pretty fast, since dividing by 3 on each step will drop the argument to arbitrarily small size in logarithmic time. Of course, don't compute sin(x/3) twice in every iteration!

Here's a proof that it works:

Thanks. I completely forgot about being able to use the "." character in \left, \right commands as you suggested. It still doesn't look great like that, but it does match what I was asking for.. lol

Now the bar definitely looks too tall though, since it matched the size of the outer brace. I guess I should have asked if it's possible to make it the next size down from the size of the brace. Does anyone have a better idea for how to do this that doesn't have the limitations this command does?

`\newcommand{\SetWithContraint}[2]{\left\{#1\,\left|\,#2\right.\right\}}`

Is there a way to do this that wouldn't be considered cheating? Sorry if it's a noob question whose correct answer is "RTFM", but I haven't seen a good example of how to do this cleanly yet. Thanks again for the help, mathstyperson!

**betterthangauss**- Replies: 5

I would like to make a command to write sets with constraints, so in my LaTeX I could write something like

`\SetWithConstraint{x}{x^2 > 0}`

Instead of having to manually do

`\left{x | x^2 > 0 \right}`

Here is the initial command definition I put in my preamble:

`\newcommand{\SetWithContraint}[2]{\left\{#1\,|\,#2\right\}}`

It seemed to work pretty well at first, for example this comes out looking pretty good:

However, I would like a way to have the size of the | in the middle match the size of the braces enclosing the set definition. My results here using that command look horrible:

Does anyone know a good way to accomplish this? I'd really prefer my TeX files weren't cluttered with garbage regarding explicit layouts of individual characters. Thanks!

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