Clarification on the proof of theorem 8 (Set Theory and Metric Spaces - Kaplansky)

Question

There are 2 parts (red underline and yellow highlight) in the proof below that I couldn’t follow.

Devise a one-to-one map of A onto C. Doesn’t the article ‘a’ mean ‘one’? However, the proof says: take g = the identity (for set D) AND take g = f (for unilateral cycles E). That means two instead of one function g, doesn’t it?
Is it a typo for the highlighted phrase (C might adjoin to A)? It should be C might adjoin to f(A) instead, shouldn’t it?

Could someone please help clarify? Thank you in advance.

Berci · Accepted Answer · 2019-11-25 00:23:34Z

1

Yes, we are constructing one function on $A$, but separately on $D$ and (two parts of) $E$.
Starting from the given example, $f(A)=D\cup\{a_2,a_3,\dots\}\cup\{b_2,b_3,\dots\}\cup \dots$, assume specifically that $C=f(A)\cup\{b_1,c_1\}$.
Then set $g(x)=x$ if $x\in D$,
$g(x)=x$ if $x\in \{b_1,b_2,\dots,\,c_1,c_2,\dots\}$,
and $g(x)=f(x)$ otherwise (so that if $x\in \{a_1,a_2,\dots,\,d_1,d_2,\dots,\dots\}$).
Yes, it is a typo.

answered Nov 24, 2019 at 23:58

Berci

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$\begingroup$ Thanks for your quick response, @Berci. If there are TWO functions g from A to C, then the proof failed to prove the point of there exists ONE function mapping between the two sets in question? $\endgroup$
– Nemo
Commented Nov 25, 2019 at 0:04
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$\begingroup$ It's just an abuse of notation. First they say $g|_D$ is an arbitrary bijection of $D$, say the identity or $f$, then they don't define $E_1$ only by words: as the elements $x$ of $E$ for which the whole $f$-chain of $x$ (and in particular, its initial element) is contained in $C$, and then $g|_{E_1}$ must be the identity, and on the rest, $E_2:=E\setminus E_1$, they write to define $g|_{E_2}$ as $f|_{E_2}$. It is one mapping. $\endgroup$
– Berci
Commented Nov 25, 2019 at 0:11
$\begingroup$ Thanks, @Berci. This makes it much clearer. One more thing: because g |E1 must be the identity, that means it doesn't count as as separate function from g |E2 (f |E2)? I've just seen your edited answer, it's so much more succinct. Thanks heaps! $\endgroup$
– Nemo
Commented Nov 25, 2019 at 0:29
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$\begingroup$ Well. Technically three functions are defined here, but note that their domains are disjoint with the union of the whole base set $A$. Since a function is set theoretically just the set of pairs $(x,f(x))$, we can calmly take the union of these 3 functions to get $g$. $\endgroup$
– Berci
Commented Nov 25, 2019 at 0:31

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