.999 = 1, in descrete topology?

[TylerH]

When one is considering a the real numbers with the extension of an infinitesimal, implying that it is possible for a number to be the highest number lower than a number, would .999... then be the highest number less than 1? (Similar to *R, but I'm generalizing my hypothesis to include any extension to the reals that fits the condition of having an infinitesimal.)

 

[Hurkyl]

When one is considering a the real numbers with the extension of an infinitesimal, implying that it is possible for a number to be the highest number lower than a number, would .999... then be the highest number less than 1? (Similar to *R, but I'm generalizing my hypothesis to include any extension to the reals that fits the condition of having an infinitesimal.)

The real number 0.999... is the real number one. If you consider other number systems, the real number 0.999... is still the real number one.


As an aside, it is impossible, in an ordered field, for there to be a largest number smaller than 1. Between any two numbers, there is always another number.


In the hyperreal numbers, the hyperreal number 0.999... (this time the notation is referring to a hyperdecimal number, rather than a decimal number) is the hyperreal number 1. What the hyperreal numbers do have is the ability to consider a number 0.999...9 with a hyperfinite (but transfinite) number of 9's, rather than having a 9 in every hyperdecimal position as in the numeral 0.999... Such a number would be infinitessimally close to 1.

 

[TylerH]

What is the difference between hyperfinite and infinite?

 

[Hurkyl]

What is the difference between hyperfinite and infinite?

They are more or less independent concepts.

Hyperfinite plays the same role in the non-standard model as finite does in the standard model. e.g. every finite set of real numbers has a largest element, and so does every hyperfinite (internal) set of hyperreal numbers. In the case of counting (hyper)decimal places, to say that 0.99...9 has hyperfinitely many 9's just says there is a hyperinteger number of 9's, and the rest of the digits are zero.

Infinite, here, has to do with comparing standard things to non-standard things -- in this case, a positive infinite hyperinteger is simply hyperinteger that isn't also a standard integer. (and, thus, is larger than all standard integers)



In the standard model, you can pick any positive integer and write down a numeral with that many 9's after the decimal place, and the rest of the digits zero. This number will be less than 1.

Transferring this to the non-standard model means you can pick any positive hyperinteger and write down a hypernumeral with that many 9's after the decimal place, and the rest of the digits zero. This number will also be less than 1.

If your hyperinteger is infinite, then the number so written will be infintiessimally close to 1.

 

[TylerH]

If your hyperinteger is infinite, then the number so written will be infintiessimally close to 1.

Is this supposed to mean they are equal? If so, why wouldn't infinitely close imply not equal?