Have you forgotten that isa true statement?

So is false.

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- May 1st 2010, 01:16 PM #1

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## Nagging question

Let and be nonempty finite sets. For to be a function, it must satisfy this condition:

.

Now, if I make and , the statement of implication becomes

, from which we know that the statement of implication is true, since

False False is a true statement. In this regard, I have a question:

Is a function?

- May 1st 2010, 02:41 PM #2

- May 1st 2010, 05:09 PM #3

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- May 2nd 2010, 11:48 AM #4

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- May 2nd 2010, 01:10 PM #5

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- May 2nd 2010, 01:33 PM #6

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- May 2nd 2010, 01:51 PM #7

- May 3rd 2010, 07:21 AM #8

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I find that unneeded commas are usually visual noise.

The notation

<0 0>

is quite clear.

I've corresponded with numerous professional and amateur mathematicians and have never had any problem with

<0 0>

being understood as the ordered pair with 0 as first coordinate and 0 as second coordinate.

/

I've already responded in another thread regarding LaTex.

- May 3rd 2010, 09:05 AM #9

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- May 3rd 2010, 09:15 AM #10

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It's just a kind of fun fact that derives from our particular set theoretic definition of the ordered pair operation:

The standard definition of the ordered pair operation:

<x y> = {{x} {x y}}

Also, we have a "redundancy property":

{x x} = {x}

Also, we have the von Neumann approach to natural numbers, in which 1 = {0}.

So:

{{1}} = {{{0}}} = {{{0} {0}}} = {{{0} {0 0}}} = {<0 0>}

And {<0 0>} is a function since it satisfies the set theoretic definition of a function:

f is a function iff (f is a set of ordered pairs & there are no members <x y> and <x z> in f unless y=z)

- May 3rd 2010, 06:15 PM #11

- May 4th 2010, 08:16 AM #12

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- May 4th 2010, 05:57 PM #13