http://i46.tinypic.com/htefd0.png

Is this sufficient? I feel like I missed something sire I did not use the S union T property.

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- Jan 22nd 2010, 02:03 PMdavismjS contains T if the intersection is T
http://i46.tinypic.com/htefd0.png

Is this sufficient? I feel like I missed something sire I did not use the S union T property. - Jan 23rd 2010, 02:21 AMtonio

What you did is enough: $\displaystyle T\subset S\Longleftrightarrow S\cap T=T$ , no need at all to assume anything about the union, BUT we also have $\displaystyle T\subset S\Longleftrightarrow S\cup T=S$ , without any assumption on the intersection, so perhaps the question was to prove that T is a subset of S iff any of those two properties, separatedly, is true.

Tonio