Re: 1-1/2+1/3-1/4+1/5-1/6+1/7

hagman wrote:

On 30 Jan., 10:05, Han de Bruijn <Han.deBru...@xxxxxxxxxxxxxx> wrote:

Virgil wrote:

In article <76e9b$479f4631$82a1e228$13...@xxxxxxxxxxxxxxxx>,
Han de Bruijn <Han.deBru...@xxxxxxxxxxxxxx> wrote:

Aatu Koskensilta wrote:

On 2008-01-29, in sci.math, Han de Bruijn wrote:

Whew! Now, _that_'s an explanation! What's the *!#!@$ difference !?

The locution "set closed under the successor function" has a perfectly
clear meaning while "applying the successor function an infinite
number of times" is meaningless.

Typical .. I'd rather say that they're exactly the same. Meaning: if you
accept the first then I see no reason why not accept the second. Talking
about "clear" ..

The set {0,1} is closed under f(x) = 1-x and also under g(x) = x^2.
Neither of these closures requires infinite numbers of compositions.

Those are _finite_ sets. I have _no_ trouble with that.

Hm, the set (0,1) is widely believed to be infinite.
However, the set IN can be bijected with a proper subset of (0,1)
via n |-> 1/(n+2), thus showing, uhm, that IN is finite?

Didn't you just mix up curly braces with common parentheses?
: {0,1} <> (0,1) .

But the naturals
are an essentially different matter.

You should explain what infinite composition of a function is,
I don't see how that requires a special domain
until you give a proper definition.

Han de Bruijn

.

Relevant Pages

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  • Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
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  • Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
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  • Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
    ... clear meaning while "applying the successor function an infinite ... Neither of these closures requires infinite numbers of compositions. ... However, the set IN can be bijected with a proper subset of ...
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