Is every (possibly infinite) sum of cardinal numbers defined?
$begingroup$
Hrbacek and Jech gives the following definition of cardinal addition:
My question is: given an indexed system of cardinals $left langle kappa_{i} |iin I right rangle$ does there exist a system $left langle A_{i} |iin I right rangle$ of mutually disjoint sets such that $|A_{i}|=kappa_{i}$ for all $i in I$?
elementary-set-theory cardinals
asked Dec 29 '18 at 2:49
Barycentric_BashBarycentric_Bash
42339
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add a comment |
$begingroup$
Hrbacek and Jech gives the following definition of cardinal addition:
My question is: given an indexed system of cardinals $left langle kappa_{i} |iin I right rangle$ does there exist a system $left langle A_{i} |iin I right rangle$ of mutually disjoint sets such that $|A_{i}|=kappa_{i}$ for all $i in I$?
elementary-set-theory cardinals
asked Dec 29 '18 at 2:49
Barycentric_BashBarycentric_Bash
42339
$endgroup$
add a comment |
$begingroup$
Hrbacek and Jech gives the following definition of cardinal addition:
My question is: given an indexed system of cardinals $left langle kappa_{i} |iin I right rangle$ does there exist a system $left langle A_{i} |iin I right rangle$ of mutually disjoint sets such that $|A_{i}|=kappa_{i}$ for all $i in I$?
elementary-set-theory cardinals
asked Dec 29 '18 at 2:49
Barycentric_BashBarycentric_Bash
42339
$endgroup$
Hrbacek and Jech gives the following definition of cardinal addition:
My question is: given an indexed system of cardinals $left langle kappa_{i} |iin I right rangle$ does there exist a system $left langle A_{i} |iin I right rangle$ of mutually disjoint sets such that $|A_{i}|=kappa_{i}$ for all $i in I$?
elementary-set-theory cardinals
elementary-set-theory cardinals
asked Dec 29 '18 at 2:49
Barycentric_BashBarycentric_Bash
42339
asked Dec 29 '18 at 2:49
Barycentric_BashBarycentric_Bash
42339
asked Dec 29 '18 at 2:49
Barycentric_BashBarycentric_Bash
42339
asked Dec 29 '18 at 2:49
Barycentric_BashBarycentric_Bash
42339
asked Dec 29 '18 at 2:49
Barycentric_BashBarycentric_Bash
42339
42339
add a comment |
add a comment |
1 Answer
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Yes. Just let $A_i={i}times kappa_i$, for instance.
answered Dec 29 '18 at 2:53
Eric WofseyEric Wofsey
190k14216348
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1 Answer
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1 Answer
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$begingroup$
Yes. Just let $A_i={i}times kappa_i$, for instance.
answered Dec 29 '18 at 2:53
Eric WofseyEric Wofsey
190k14216348
$endgroup$
add a comment |
$begingroup$
Yes. Just let $A_i={i}times kappa_i$, for instance.
answered Dec 29 '18 at 2:53
Eric WofseyEric Wofsey
190k14216348
$endgroup$
add a comment |
$begingroup$
Yes. Just let $A_i={i}times kappa_i$, for instance.
answered Dec 29 '18 at 2:53
Eric WofseyEric Wofsey
190k14216348
$endgroup$
Yes. Just let $A_i={i}times kappa_i$, for instance.
answered Dec 29 '18 at 2:53
Eric WofseyEric Wofsey
190k14216348
answered Dec 29 '18 at 2:53
Eric WofseyEric Wofsey
190k14216348
answered Dec 29 '18 at 2:53
Eric WofseyEric Wofsey
190k14216348
answered Dec 29 '18 at 2:53
Eric WofseyEric Wofsey
190k14216348
190k14216348
add a comment |
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