Let $K$ be a field and $V$ a $K$-vector space and $U_1, U_2 \subseteq V$ subspaces of $V$ that satisfy $U_1 \not\subseteq U_2$ and $U_2 \not\subseteq U_1$. Prove that $U_1 \cup U_2$ is not a subspace of $V$.

I only found the contraposition of this argument here and here.

My counterexample goes as follows:

One obtaines $U_1 \not\subseteq U_2$ and $U_2 \not\subseteq U_1$ $\implies U_1 \cap U_2 = \emptyset$. For $U_1 \cup U_2$ to be a subspace of $V$ we have to prove the three subspace axioms, the first being that it's not empty.

But if we choose $$ V := \{ p \in \mathbb{R}[t]: deg(p) < 2 \}, \quad U_1 := \{ p \in V: p(0) = 0 \}, \text{ and } U_2 := \{ p \in V: p(0) = 1 \} $$ We obtain $U_1 \cap U_2 = \emptyset$.

Is my counterexample correct?

  • 3
    $\begingroup$ In the example, $U_2$ is not a subspace : do you think it contains zero, or is closed under addition? Two subspaces cannot have empty intersection , since zero is in the intersection. $\endgroup$ – астон вілла олоф мэллбэрг Aug 28 '18 at 15:01

Hint for the proof: Choose $u_1\in U_1\backslash U_2$ and $u_2\in U_2\backslash U_1$ prove that $u_1+u_2\not\in U_1\cup U_2$.

If by contradiction $u_1+u_2$ is an element of, say $U_1$ what can you say about $u_2$?


One obtains $U_1 \not\subseteq U_2$ and $U_2 \not\subseteq U_1$ $\implies U_1 \cap U_2 = \emptyset$.

No. For one thing $0\in U_1\cap U_2$ always.

But if we choose et cetera.

$U_2$ is not a vector subspace, because $0\notin U_2$.

In fact your (like any other) counterexample cannot be correct, because the theorem is true.

  • $\begingroup$ I got it know, the statement is incorrect because $U_1 \not\subseteq U_2$ doesn't rule out that $U_1$ and $U_2$ share elements. $\endgroup$ – Viktor Glombik Aug 28 '18 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.