True or false: the empty set is convex. prove your claim.

What are your thoughts? What definitions or theorems have you applied? How far have you gotten? Where are you stuck?

Please be complete. Thank you! :D

Eliz.
 
If \(\displaystyle \left\{ {x,y} \right\} \subseteq \emptyset\) then \(\displaystyle \left( {\forall t \in \Re } \right)\left[ {\left( {tx + \left( {1 - t} \right)y} \right) \in \emptyset } \right]\).
That is a true statement( recall that a false statement implies any statement).
Thus that is a valid proof.
 
So, even though no such x,y can exist in the empty set, the statement is true? And that proves that the empty set is convex because a set is convex if tx + (1-t)y?
 
The set \(\displaystyle C\) is convex if \(\displaystyle \left\{ {x,y} \right\} \subseteq C\) then for \(\displaystyle 0 \le t \le 1\) we have \(\displaystyle \left( {tx + (1 - t)y} \right) \in C\).

The empty-set fills that definition.
 
i am not convinced

my tutor says the empty set is not convex.
to prove this he says:
first define the empty set
then define convex set.
by definition, the convex set is not empty
i.e. we proved that there is
tx + (1-t)y element of A, whereas the empty set has not elements
therefore, the empty set is..."not a subset of" symbol...convex set.
I tried to write out his mathy notation...does it make sense?
 
Re: i am not convinced

woolley said:
my tutor says the empty set is not convex. to prove this he says:
first define the empty set, then define convex set.
by definition, the convex set is not empty
Click to expand...
I do not want to contradict your tutor. Perhaps he uses a different definition of convex set.
However, if one defines a convex set as being nonempty, then why ask if the empty set is convex?
That is what does not make any sense.
 
semantics

I see the confusion. My instructor agrees with you. I guess I should provide my tutor with the definitions we are using in class.

Thanks!
 
You must log in or register to reply here.
Top