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Is the set \(\displaystyle{\left(0,1\right)}\) an algebraic subset of \(\displaystyle{\mathbb{A}^{1}_{\mathbb{R}}}\) ?

The answer is negative.

Proof

Suppose that the set \(\displaystyle{\left(0,1\right)}\) is an algebraic subset of \(\displaystyle{\mathbb{A}_{\mathbb{R}}^{1}}\) .

Then, \(\displaystyle{\left(0,1\right)=V(f(x))}\), where \(\displaystyle{f(x)\in\mathbb{R}[x]}\).

But, according to the Fundamental Theorem of Algebra, \(\displaystyle{V(f(x))}\) is finite, a contradiction,

since, \(\displaystyle{\left(0,1\right)}\) is infinite.