1. Originally Posted by SkyWatcher
for problem two i think it is useless to proove anything about U S and T when they are the empty set because they have no element so we cannot say anything about the product of two element of such set.
usualy set closed by order are groups (which are none empty) or that sort of thing
beter look to the definition of a set closedby multiplication to see if it admits empty set by definition but demonstration is useless about what do or do not elements of the empy set!
What are you saying? The sets $T,S$ are not necessary empty. For example, take $U = \mathbb{R} - \{ 0 \}$. And take $T= \mathbb{R}^+ \mbox{ and } S = \mathbb{R}^-$. Their intersection is empty and union is $U$. Furthermore, they obey the property that the product of any three elements is again in the set!

Also, an empty set is closed. Because the condition is $\mbox{ If }a,b \in S \mbox{ then }ab\in S$. This is a true statement. See Here.

2. i was only talking about your first sentence of number two I assumed that when you wrote {} you where speaking about the empty set and nothing else you said the proof is imediate
i said elsewhere recently that you can proof anything you want about the element of the empty set and the contrary
i agree when you say if a and b belongs to S (the empty set) it implyes that a*b belongs to S this is a true statement because they are no a and b belonging to S
but you can proof the same way that s (the empty set) is unclosed a*b would not belong to S even if a and b where both belonging to S

3. Originally Posted by SkyWatcher
but you can proof the same way that s (the empty set) is unclosed a*b would not belong to S even if a and b where both belonging to S
No you cannot!

4. yes your right i focussed that point of beeing aware of the statement you can deduce by considering things that never hapen wich is explained unfortunatly in english in the 'see there' above for an other reason of being purely finicaty
i didnt see the IF that is on the definition
but in reality i was used to deal whith that closing or unclosing matter with groups or dual and i doubt (that is my pericularity) that an empty set can be a group or a dual (maybe you don't know the word group) because of the abcence of some neutral elements

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