kuratowskis theorem (a,b)={{a},{a,b}}

Onyx · 2009-03-10 10:14:37

I have this question:

Let (a,b) denote {{a},{a,b}} then which of the following are true:

a.)

TRUE

b.)

FALSE

c.)

TRUE

d.)

TRUE

e.)

FALSE

f.)

FALSE

I'm not sure I understand the theorem, but I think my answers are right. Can someone check?

JaneFairfax · 2009-03-10 10:51:51

You goit (c) and (e) correct. Your other answers are incorrect, unfortunately.

(a) is false
(b) is true
(d) is false
(f) is true.

JaneFairfax · 2009-03-10 11:12:30

Wait a minute. Your answer to (c) is incorrect as well.

So (c) is false.

My, that was tricky. Even I got fooled.

TheDude · 2009-03-10 23:58:13

You're not using their definition. Whenever you see numbers in parentheses you need to mentally replace them by the set. For example, the first question asks if

. To answer the question you need to replace (a,b) with {{a},{a,b}}. You can see that the answer is actually false now (remember like Jane said that ).

Onyx · 2009-03-11 13:38:34

Thanks for your replies, what you've said is initially what I thought, in which case I would have go them all, but then I changed my mind because of this reasoning:

JaneFairfax · 2009-03-11 22:01:33

Onyx wrote:

JaneFairfax · 2009-03-11 22:06:40

Onyx wrote:

Onyx · 2009-03-11 23:29:25

Yes I know that, but since {a} is a set and {{a},{a,b}} is a set, and since a is an element of a set, I would have thought I used the relations correctly.

P.S. keep your hair on jane!

Last edited by Onyx (2009-03-11 23:30:37)

TheDude · 2009-03-12 00:02:41

Onyx wrote:

Yes I know that, but since {a} is a set and {{a},{a,b}} is a set, and since a is an element of a set, I would have thought I used the relations correctly.

Remember that anything can be an element of a set, including other sets. In the case of {{a}, {a, b}} every element of this set is another set. Since a by itself is not a set it cannot be an element of {{a}, {a, b}}.

Onyx · 2009-03-12 04:41:53

Ok I get it now. my reasoning doesn't work since

and I see what you were saying.

Thanks

Onyx · 2009-03-12 23:17:40

Hi, I've got another question now regarding kurwatowskis theorem. I need to prove:

It seems like it should be so obvious, but I'm having trouble writing a meaningful proof for it, without actually just replacing a with u and b with v.

I'm given the hint that I need to consider the two cases where a=b and

My thoughts are that these cases are:

since a=u and b=v,

I feel like I'm making a mess of it, please help!

LuisRodg · 2009-03-14 07:16:40

Onyx · 2009-03-15 10:16:05

Thanks alot thats a great help.

Math Is Fun Forum

#1 2009-03-10 10:14:37

kuratowskis theorem (a,b)={{a},{a,b}}

#2 2009-03-10 10:51:51

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#3 2009-03-10 11:12:30

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#4 2009-03-10 23:58:13

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#5 2009-03-11 13:38:34

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#6 2009-03-11 22:01:33

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#7 2009-03-11 22:06:40

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#8 2009-03-11 23:29:25

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#9 2009-03-12 00:02:41

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#10 2009-03-12 04:41:53

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#11 2009-03-12 23:17:40

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#12 2009-03-14 07:16:40

Re: kuratowskis theorem (a,b)={{a},{a,b}}

#13 2009-03-15 10:16:05

Re: kuratowskis theorem (a,b)={{a},{a,b}}

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