jkm89 said:So for my homework I have to prove (or disprove) this statement:
If U, V are two subspaces of Rn then U [tex]\cap[/tex] V [tex]\neq[/tex] [tex]\phi[/tex].
I just want to make sure; [tex]\phi[/tex] is the null set right? The set with nothing in it?
rasmhop said:Compare
[tex]\phi \quad \emptyset[/tex]
The first is phi and the second is "empty set".
The intersection of two sets U and V in the n-dimensional space, denoted as U ∩ V, refers to the set of all elements that are common to both U and V. In other words, it is the set of all points that are contained in both U and V.
In mathematics, U and V intersection in Rn is represented using set notation as U ∩ V = {x ∈ Rn | x ∈ U and x ∈ V}. This means that the intersection of U and V in Rn is a set containing all the elements x that are both in U and V.
The intersection of two sets in Rn is a fundamental concept in mathematics and has many applications in various fields such as geometry, linear algebra, and topology. It helps us understand the relationship between sets and allows us to solve problems involving multiple sets.
To prove that the intersection of U and V in Rn is non-empty, we can use the fact that if two sets have a common element, then their intersection will also have that element. Therefore, we need to show that there exists at least one element that is present in both U and V.
Yes, it is possible for the intersection of U and V in Rn to be an empty set. This happens when the two sets do not have any common elements, meaning that there is no point in the n-dimensional space that is present in both U and V. In this case, the intersection would be represented as U ∩ V = ∅, where ∅ denotes an empty set.