Subset vs. Proper Subset

[TheWrathOfMath]

If the subspaces V,U,W ⊂ R^6, and dimV=dimW=5, dimU=3, then U⋂V⋂W={0}

I know that the answer is no, and I can take U ⊆ V=W, but I managed to find an example for V,U,W ⊆ R^6 instead of V,U,W ⊂ R^6.

W=V={(1 0 0 0 1 0) (0 1 0 0 0 0) (0 0 1 0 0 0) (0 0 0 1 0 0) (0 0 0 0 0 1) (1 1 1 1 1 1)}
dimU=dimV=5

U= {(1 1 0 0 0 0) (0 0 1 1 0 0) (0 0 0 0 1 1)}
dimU=3

U⋂V⋂W =/= {0}

The problem is that, as stated above, V,U,W ⊆ R^6 instead of V,U,W ⊂ R^6.

How do I fix this?

 

[blamocur]

Either I am missing something, or there is no difference between [imath]\subset[/imath] and [imath]\subseteq[/imath], meaning that it can only be [imath]\subset[/imath] since in your case dim(L)<6. I.e., if the dimension is less than that of the enclosing space than it is always proper subset, and if the dimensions are the same than you have equality, i.e. if [imath]\dim(V) = 6[/imath] then [imath]V = \mathbb{R}^6[/imath]

 

[Steven G]

??????????

 

[TheWrathOfMath]

Either I am missing something, or there is no difference between [imath]\subset[/imath] and [imath]\subseteq[/imath], meaning that it can only be [imath]\subset[/imath] since in your case dim(L)<6. I.e., if the dimension is less than that of the enclosing space than it is always proper subset, and if the dimensions are the same than you have equality, i.e. if [imath]\dim(V) = 6[/imath] then [imath]V = \mathbb{R}^6[/imath]

I got you.
So the counterexample I provided works, I believe.
Thank you for the clarification.

 

[TheWrathOfMath]

??????????

Perhaps a better question to ask is "does the counterexample I provided works or not?"

Claim: If the subspaces V,U,W ⊂ R^6, and dimV=dimW=5, dimU=3, then U⋂V⋂W={0}

False.

Counterexample:

W=V={(1 0 0 0 1 0) (0 1 0 0 0 0) (0 0 1 0 0 0) (0 0 0 1 0 0) (0 0 0 0 0 1) (1 1 1 1 1 1)}
dimU=dimV=5

U= {(1 1 0 0 0 0) (0 0 1 1 0 0) (0 0 0 0 1 1)}
dimU=3

U⋂V⋂W =/= {0}

 

[blamocur]

You can show that the conjecture is false without using counterexamples. In fact, while counterexamples show that it is not always true, you can prove that it is never true:

[math]\dim (U\cap V) \geq \dim(U) + \dim(V) - 6 = 4[/math]
[math]\dim (U\cap V\cap W) = \dim ((U\cap V) \cap W) \geq \dim(U\cap V) + \dim(W) -6 = \dim((U\cap V) +3 - 6 \geq 4+3-6 = 1[/math]
Subspace of dimension at least one cannot be zero.