Thread: 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?

Have you forgotten that is a true statement?
So is false.

Originally Posted by Plato

Have you forgotten that is a true statement?
So is false.

In deed, I forgot about it. Oh, Sir, I had a brain fade at noon right after eating a full meal.

So we see:

0 is a function from 0 into 0

By the way, another cute one is this:

[EDITED OUT; CORRECTED BELOW]

Originally Posted by MoeBlee

So we see:

0 is a function from 0 into 0

By the way, another cute one is this:

{1} is a function

{1} = {{0}} = {{0} {0 0}} = {<0 0>} is a function

No comprehende, amigo!

Originally Posted by novice

No comprehende

That's because I made a mistake.

I meant to say {1} is an ordered pair and {{1}} is a function.

{{1}} = {{{0}}} = {{{0} {0 0}}} = {<0 0>} is a function

Originally Posted by MoeBlee

That's because I made a mistake.

I meant to say {1} is an ordered pair and {{1}} is a function.

{{1}} = {{{0}}} = {{{0}, {0, 0}}} = {<0, 0>} is a function

MoeBlee
I am curious at why you do not use commas in ordered pairs as is the standard notation.
Why do you use in place of the standard ?
You know that can confuse many who use resource.

Again, why don’t you learn to use LaTeX?

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.

Originally Posted by MoeBlee

That's because I made a mistake.

I meant to say {1} is an ordered pair and {{1}} is a function.

{{1}} = {{{0}}} = {{{0} {0 0}}} = {<0 0>} is a function

I must thank you for your willingness in helping me with my learning, though I have not learned enough mathematics to understand what is being imparted to me. Hopefully, the day will come when I can recognize it and appreciate it. Mucho Gracias.

Originally Posted by novice

I have not learned enough mathematics to understand what is being imparted to me.

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)

Originally Posted by MoeBlee

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.

That may well be the case. Nevertheless, your notation is not standard notation and has the potential to create confusion. I support Plato's remarks and request that you follow them when making future posts.

P.S. Actually, the notation

(x, y)

is more ambiguous than

<x y>.

(x, y)

could be taken as the ordered pair of x and y, or as an open real interval.

But

<x y>

is quite clearly the ordered pair of x and y.

Originally Posted by MoeBlee

P.S. Actually, the notation

(x, y)

is more ambiguous than

<x y>.

(x, y)

could be taken as the ordered pair of x and y, or as an open real interval.

But

<x y>

is quite clearly the ordered pair of x and y.

This thread is taking a direction beyond the question asked by the OP.

Thread closed.