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So I had a test today, and one of the questions was:

"Give an example of a set, S contained in $R^3$ that is closed under scalar multiplication but isn't closed under addition"

And I was just wondering -- is there a systematic way to go about this rather than search for examples? Perhaps some sort of algebraic way to create such a set that satisfies these conditions without any further thought?

Hugh Entwistle
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  • There is no good, general way to come up with counterexamples "via a procedure, without any further thought". That being said, it helps to have a few such examples in the back of your mind. One important example, as seen here, is two vector subspaces whose union fails to be a vector space. – Ben Grossmann Oct 12 '16 at 13:19

2 Answers2

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Finding counterexamples is always specific to the kind of problem you have at your hand. Particular properties must be exploited.

Here, closedness under scalar multiplication precisely means that if vector is contained in our set $S$, then the whole line through origin containing the vector is. So, our set $S$ must have at least one line. This wouldn't be counterexample, since lines through origin are additively closed, so at least one more line must be added to $S$. Now, we have two lines through origin in $S$. For this set to be additively closed would mean that $S$ contains the whole plane determined by our lines (you have two independent vectors in $S$). So, for our counterexample to work, we must have at least two lines in $S$, but not the whole plane spanned by any of the two. You could, for example pick coordinate axes in $\mathbb R^3$, or anything else of the sort that tickles your fancy.

In more general terms, you have two subspaces $U,V\subseteq X$. Then their union is never additively closed, unless one is contained in the other. That is why we define $U+V$ as minimal subspace that contains both $U$ and $V$ - we can't just take $U\cup V$.

Ennar
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Let $u = \left( u_1, u_2, u_3 \right)$ and $v = \left( v_1, v_2, v_3 \right)$ be any two elements of $\mathbb{R}^3$ such that $$\frac{u_1}{v_1} = \frac{u_2}{v_2} = \frac{u_3}{v_3}$$ does not hold.

Now let $U$ denote the set of all the scalar multiples of $U$, and let $V$ denote the set of all the scalar multiples of $v$. Then $U \cup V$ is your desired set.

Saaqib Mahmood
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