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Could someone lend me assistance by explaining why

a semi-algebraic set has dimension zero $\Leftrightarrow$ it is finite or empty

is true or by giving a reference?

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up vote 4 down vote accepted

You can find it all in Bochnak,Coste,Roy: Real Algebraic Geometry:

(1) up to semi-algebraic homeomorphism, every semi-algebraic subset is a disjoint union of finitely many (open) hypercubes (i.e. sets of the form $]0,1[^d$ for some $d\in\mathbf{N}$). This is Theorem 2.3.6

(2) a nonempty open semi-algebraic subset of $R^n$ has dimension $n$. This is Proposition 2.8.4

(3) the dimension of a finite union of semialgebraic sets $A_1,\ldots,A_p$ is $\max(\dim(A_1),\ldots,\dim(A_p))$. This is Proposition 2.8.5(i)

(4) dimension is invariant under semialgebraic homeomorphism. This is included in Theorem 2.8.8

Therefore, if a semi-algebraic set is zero-dimensional, its decompositon as in (1) can only be made up from hypercubes of the form $]0,1[^0$, i.e. points.

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Wikipedia says that "it is not hard to see that a semialgebraic set lies inside an algebraic subvariety of the same dimension". That should suffice.

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