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Closed set: If the result of a given operation including two or more members of a set is in the set, it is considered closed. Therefore, wouldn't you need to include a given operation with your question? The set of reals is closed under addition and subtraction, but open under division, so simply summarizing its closure in general cannot be done.

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Is PJ

Well, if we use the definition given above:

Quote (Caesar)

Closed set: If the result of a given operation including two or more members of a set is in the set, it is considered closed.

Then the null set (i.e., '{}') is closed. Within the set of all real numbers, one can divide by zero, and the result is not within the set of real numbers. Therefore, the set of all real numbers is open.

Here, by 'operation', I assume the standard +,-,*,/.

If I'm correct, question goes to Caesar, because he provided the meat and potatoes of the answer.

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Check the "What games are you playing at the moment?" thread for updates on what I've been playing.

You can find me on the Fediverse! I use Mastodon, where I am @[email protected] ( https://sakurajima.moe/@glennmagusharvey )

i'm not sure where you got that definition for "closed set." there are two outstanding theoretical definitions that come to mind, and neither is close to the definition you have presented. furthermore, if i'm understanding your definition correctly, it's not even true. let's make an arbitrary closed set of three integers: {1,2,3}. i doubt you'd argue that this is an open set, but let's see if not by your definition we come across that result. for example, let's ask ourselves if the set is open or closed under addition. taking 3 and 2, which are both members of our set, and adding them to one another, we get 5, which is *not* in the set. therefore, our set is, by your definition, "open under addition."

i will solve the riddle later today if no one has posted a solution.

Quote (gozaru~ @ 9th May 2005 13:36)

i'm not sure where you got that definition for "closed set." there are two outstanding theoretical definitions that come to mind, and neither is close to the definition you have presented. furthermore, if i'm understanding your definition correctly, it's not even true. let's make an arbitrary closed set of three integers: {1,2,3}. i doubt you'd argue that this is an open set, but let's see if not by your definition we come across that result. for example, let's ask ourselves if the set is open or closed under addition. taking 3 and 2, which are both members of our set, and adding them to one another, we get 5, which is *not* in the set. therefore, our set is, by your definition, "open under addition."

i will solve the riddle later today if no one has posted a solution.

I see not the purpose of your example. The set {1, 2, 3} is open under addition, as my definition declared. I'm not sure if you were trying to refute my definition with that example or what exactly.

Either way, as per all trivia threads I respond in, I refuse to look up the solution and take credit for it, so I stick to my previous answer, since that's all that comes to mind off the top of my head.

Btw, that defintion was composed spur-of-the-moment from what knowledge I retained from learning it in Math class last year (or whatever year it was).

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Is PJ

ok, well, that's all great and wonderful, but no, the set {1,2,3} is *absolutely* a closed set. i *think* i've grasped your misunderstanding: you're assuming that a given set, to be determined either open or closed, must also be associated with some operational condition. that is, you seek to say that we can only define the openness of sets in terms of the elementary functions.

...

what we COULD say is that, over R, the set {1,2,3} is open under addition on some interval that includes all possible sums of elements of the set. for example, we could say that {1,2,3} is open under addition over R or (0,100) or any arbitrary interval that we wanted to use that satisfies the condition. but of course we do not say thusly that {1,2,3} is an "open set." you're confusing two entirely different concepts.

anyway, to move away from the botched defintion of open sets that has been presented, let's describe an open set as any set for which all members have an epsilon-neighbourhood not extending outside the boundaries of the set. i.e.,

set "s" is open iff for all members "x" belonging to s there is an open interval (or any of its higher-dimensional analogues) inside s that contains x.

for example: (1,5), or {x:1<x<5}, is open. no matter how close you get to 1, you can always get even closer without equalling 1. [1,5], or {x:1<=x<=5}, is closed. since the point 1 lies this time within the set, to get any further from 1 in the negative direction would be to extend out of the set. so, in this case, we cannot draw an interval around 1 that is still enclosed within the set.

visually,

|------(-------|-------)---------------/\/----|
1----9E-5--1E-4--1.1E-4-------------------5

the definition of closed set is a bit simpler: a set is closed iff its complement is open; i.e., a set "s" is closed iff R - s is open. another way to express this is by saying that the set contains all of its limit points.

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ANSWER HERE\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

so the sets R and the null set are both open and closed at the same time!

/////////////////////////////////////eoa///////////////////////////////////

why? well, let's take R, to begin. for all numbers in R, an open interval can be drawn that does not exceed positive or negative infinity. that is, one can always get closer to infinity without ever reaching it. so, R is open. but, at the same time, R contains all of its limit points: on R, for every member x there is a y which also exists on R such that |y-x| < any epsilon > 0. this is a bit unwieldy, so in anticipation of our discovering that the null set is open, we can say hence that, since the complement of R, the null, is open, then R is closed.

the null set is the interval from 0 to 0 not including 0. thus, for every member of the null set (there are no members), it is possible to find an epsilon-neighbourhood not extending outside the boundaries. therefore, the null set is open. but wait! we just said null's comlement was open. therefore, null must be closed. similarly, the limit points of the null set -- for which there are none -- are all enclosed in the null set, which is completely empty.

for those interested, i believe both sets are normally described as being open sets.

Hmm...yea that's a different definition of closed and open, I'd say. My definition for closed set is accurate, however, with what we were doing, and so I assumed open was simply the opposite. My bad.

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Is PJ

It's been four months, so lets revive this.

A fortress lies in a cold light
With silky walls of marble white
Their are no doors to this stronghold
Yet fools break in to steal the gold

*HINT* The first line makes no sense, except for keeping the rhyme. Besides that, its the best riddle I can think of

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Dead Winter. A nice comic.

could this be an egg?

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"Fire and steel
follow me through the lands
you will burn hordes of hell
in the deadly raging flames of revenge"

Rhapsody - Flames of Revenge

Yep, Alpha... sounds like an egg.

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Join the Army, see the world, meet interesting people - and kill them.

~Pacifist Badge, 1978

Sorry for being gone so long. Screwed up internet on my computer. Yes it is an egg.

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Dead Winter. A nice comic.

I heard this today and wanted to post it, and, wow, a riddle thread. I know it's not my turn, but it seems better to jump in here than to start a new thread just for me.

Sir, I bear a rhyme excelling
In mystic force and magic spelling
Celestial sprites elucidate
All my own striving can't relate

This one needs a hint: pay attention to the individual words even more so than the meaning of each line, especially since the meaning of the lines are fairly archaic.

And here's another hint: math.

Moderator Edit

Anyone getting this does not get a turn because it's not Kara's to give, unless Alphasmart gives up his turn. -R51

This post has been edited by Rangers51 on 30th September 2005 18:36

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Veni, vidi, dormivi.