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- Newsgroups: sci.math
- Path: sparky!uunet!mcsun!fuug!funic!nokia.fi!newshost!chalcraft
- From: chalcraft@uk.tele.nokia.fi (Adam Chalcraft)
- Subject: Re: Partitioning of uncountable sets
- In-Reply-To: kevin@vaxc.cc.monash.edu.au's message of 8 Sep 92 18:27:06 +1000
- Message-ID: <CHALCRAFT.92Sep8131748@laurel.uk.tele.nokia.fi>
- Sender: usenet@noknic.nokia.fi (USENET at noknic)
- Nntp-Posting-Host: laurel.uk.tele.nokia.fi
- Organization: cpd
- References: <1992Sep8.182706.90039@vaxc.cc.monash.edu.au>
- Date: Tue, 8 Sep 1992 11:17:48 GMT
- Lines: 22
-
- In article <1992Sep8.182706.90039@vaxc.cc.monash.edu.au> kevin@vaxc.cc.monash.edu.au writes:
-
- > Let X be an uncountable set. Consider the set of ordered pairs, (x,0),(x,1)
- > where x is in X. Call this set Y. Then Y is also uncountable, moreover,
- > {(x,0) with x in X}=Y(0) and {(x,1) with x in X}=Y(1) are both uncountable.
- > But the cardinality of X is the cardinality of Y.
-
- > The proof is of course easier if one assumes A.C, which I have avoided.
-
- I'm probably being particularly dense here, but why (without AC) do X and Y
- have the same cardinality? (Two sets have the same cardinality iff there is
- a bijection between them, of course).
- --
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