Example of a closed set under scalar multiplication that is not a subspace

Expert intro: This question tests a subtle subspace criterion: closure under scalar multiplication alone is not enough to guarantee a subspace.

Detailed walkthrough

A valid example

Let $V=\mathbb{R}^2$ over the scalar field $\mathbb{R}$, and define

$U=\{(x,0): x\in\mathbb{R}\} \cup \{(0,y): y\in\mathbb{R}\}.$

This is the union of the two coordinate axes.

It is a proper non-empty subset of $V$, and it is closed under scalar multiplication: if $u\in U$ and $k\in\mathbb{R}$, then $ku$ is still on the same axis as $u$, so $ku\in U$.

However, $U$ is not a subspace of $V$ because it is not closed under addition. For example,

$(1,0)\in U \quad \text{and} \quad (0,1)\in U,$

but

$(1,0)+(0,1)=(1,1)\notin U.$

So $U$ satisfies scalar-multiplication closure but fails the addition axiom, which means it is not a subspace.

Why this works

A subspace must satisfy three core conditions: it must contain the zero vector, be closed under addition, and be closed under scalar multiplication. Many students remember only the scalar-multiplication test and forget that addition is equally necessary.

In this example, $U$ contains $(0,0)$ because the origin lies on both axes. It is also closed under scalar multiplication. The failure happens only when vectors from different axes are added together.

General principle

A proper subset can be closed under scaling without being a subspace if it is made from several “pieces” that are individually stable under scalar multiplication but do not stay closed when combined by addition. The union of subspaces is a common source of such examples, provided the union is not itself a subspace.

Evidence and explanation

To verify the claim rigorously, we check each requested property:

1. Proper subset: $U \neq V$ because, for instance, $(1,1)\in V$ but not in $U$.
2. Non-empty: $(0,0)\in U$.
3. Closed under scalar multiplication: if $(x,0)\in U$, then $k(x,0)=(kx,0)\in U$; if $(0,y)\in U$, then $k(0,y)=(0,ky)\in U$.
4. Not a subspace: addition fails, as shown above.

That satisfies the requirement exactly.

💡 Pitfall guide

A frequent mistake is to give a set that is not even closed under scalar multiplication, such as a single open disk or a half-plane. That does not answer the question. Another error is to choose a subset that contains only the zero vector, because that is actually a subspace. The key is to find a proper non-empty set that survives scaling but breaks under addition. The union of two axes is a clean example because the failure is easy to demonstrate with a single counterexample.

🔄 Real-world variant

If the scalar field were changed to $\mathbb{C}$ and the subset were the union of two lines through the origin in $\mathbb{C}^2$, the same idea would still work as long as the lines are distinct and not closed under addition together. Another useful variant is $U=\{(x,0):x\in\mathbb{R}\}\cup\{(x,x):x\in\mathbb{R}\}$ in $\mathbb{R}^2$; it is still closed under scalar multiplication but not under addition, since adding one vector from each line can produce a vector on neither line.

🔍 Related terms

subspace criterion, closure under scalar multiplication, closure under addition