0.9999... != 1

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Originally posted by drnihili

There may be cardinals in between \aleph_0 and it's powerset, but the powerset is always bigger.
oh, yes, i think this is what is known as the continuum hypothesis. it was shown to be independent of ZFC, but is usually taken to be false, if i recall correctly.


So, with infinite cardinals, X+1=X. In fact X+X=X.
yes. so the mapping X |--> X+1 is not injective, which is a requirement for the succesor map in peano s axioms. is the injectivity of the successor map required to prove the induction principle? i wouldn t be surprised.



\omega and \omega+1 have the same cardinality, but different ordinality. Ordinal arithmetic is importantly different from cardinal arithmetic in this respect.
so then the successor map on the ordinals is injective?

An important point is that \omega is not the successor of any other ordinal. This follows from the fact that it has no last member. This is why induction whithout limit steps cannot prove anything about infinite. If you prove that 0 has a property, and that if a number has the property, so does it's successor, it's still an open question whether \omega has the property. You have to explicitly prove that limit ordinals have the property.
i see. proof by weak induction applies only to the range of the successor map. anything that is not in that range (like 0 and \omega) have to be done seperately and explicitly.



But in this case you're now doing indcution on transfinite ordinals and you have to include the limit step.
so what exactly do you mean when you say "limit step"? i am guessing that it is that stronger statement of the strong induction principle? i.e. that your statement is true for all numbers less than n implies it s also true for n, which would be explicitly required if n is not the successor of anything. is this correct?

*hopes that was relatively clear*
yeah, that was pretty good. thanks.
 
Originally posted by lethe
yes. so the mapping X |--> X+1 is not injective, which is a requirement for the succesor map in peano s axioms. is the injectivity of the successor map required to prove the induction principle? i wouldn t be surprised.


so then the successor map on the ordinals is injective?

Taking +1 to be the successor function, yes. Typically it's just taken to mean that +1 does not define a successor function for cardinals.

*edit* rereading your comment more carefully, a simple yes would have sufficed as an answer. I'm not sure about the proof of the induction pricniple, but I suspect you're right.



i see. proof by weak induction applies only to the range of the successor map. anything that is not in that range (like 0 and \omega) have to be done seperately and explicitly.



so what exactly do you mean when you say "limit step"? i am guessing that it is that stronger statement of the strong induction principle? i.e. that your statement is true for all numbers less than n implies it s also true for n, which would be explicitly required if n is not the successor of anything. is this correct?


yeah, that was pretty good. thanks.

The limit step is just the explicit proof that all "limit ordinals" have the property of the induction. It's not exactly the same as the strong inductive step, but it's similar. It's not the same because the strong inductive step has to prove the property for all least upper bounds, whereas the limit step in weak induction only has to prove it for least upper bounds which are not successors. In practice there is often a difference in how difficult these are, although they sound very much alike.
 
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Originally posted by drnihili

The limit step is just the explicit proof that all "limit ordinals" have the property of the induction. It's not exactly the same as the strong inductive step, but it's similar. It's not the same because the strong inductive step has to prove the property for all least upper bounds, whereas the limit step in weak induction only has to prove it for least upper bounds which are not successors. In practice there is often a difference in how difficult these are, although they sound very much alike.

and by limit ordinals, you just mean those ordinals that are not the successor to any other number?
 
.

are you serious? do you think it's just a game, if 0.9999999999999...=1, then the univerce, and everything, is finished, doomed, vanished, or whatever you call it, cause in that case, no balance,

so
simple answer
no!!
1 doesnt ecaul 0.999999999999999999999999999999999...
1 doesnt ecaul to 1.00000000000000000000000000000000000000000000000000000000...0000000001
 
oh, yes, i think this is what is known as the continuum hypothesis. it was shown to be independent of ZFC, but is usually taken to be false, if i recall correctly.


yes. so the mapping X |--> X+1 is not injective, which is a requirement for the succesor map in peano s axioms. is the injectivity of the successor map required to prove the induction principle? i wouldn t be surprised.


so then the successor map on the ordinals is injective?

i see. proof by weak induction applies only to the range of the successor map. anything that is not in that range (like 0 and \omega) have to be done seperately and explicitly.



so what exactly do you mean when you say "limit step"? i am guessing that it is that stronger statement of the strong induction principle? i.e. that your statement is true for all numbers less than n implies it s also true for n, which would be explicitly required if n is not the successor of anything. is this correct?


yeah, that was pretty good. thanks.

oh, yes, i think this is what is known as the continuum hypothesis. it was shown to be independent of ZFC, but is usually taken to be false, if i recall correctly.

Wrong, it is used extensively in Model Theory.
 
necromancerM.jpg
 
so
simple answer
no!!
1 doesnt ecaul 0.999999999999999999999999999999999...

I think taking math advice from someone who spells the word ``ecual'' with a c is about the last thing anyone should do.

You're an idiot.

0.99... = 1.

Period.

Thread closed.

Precedent be damned.
 
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