Answer to this question provides an algebraically independent set of real numbers of size continuum. Enumerate it as $\alpha_{i,r}$ where $i\in\{1,...,d\},r\in\Bbb R$. Now define set $S=\{(\alpha_{1,r},...,\alpha_{d,r}):r\in\Bbb R\}\subseteq\Bbb R^d$. Now, measure of any simplex with vertices on these points can be expressed using coordinates of the vertices and radicals. If any of these turned out to be an algebraic number, we would have a non-trivial algebraic relation between $\alpha_{i,r}$, which is not possible, as they are algebraically independent.

As $|S|=\frak{c}$, we see that transcendental measure set can have size $\frak{c}$. By taking subsets of $S$, we can get transcendental measure set of any size at most continuum, and, clearly, not of any greater cardinality.

Answer to this question provides an algebraically independent set of real numbers of size continuum. Enumerate it as $\alpha_{i,r}$ where $i\in\{1,...,d\},r\in\Bbb R$. Now define set $S=\{(\alpha_{1,r},...,\alpha_{d,r}):r\in\Bbb R\}\subseteq\Bbb R^d$. Now, measure of any simplex with vertices on these points can be expressed using coordinates of the vertices and radicals. If any of these turned out to be an algebraic number, we would have a non-trivial algebraic relation between $\alpha_{i,r}$, which is not possible, as they are algebraically independent.

As $|S|=\frak{c}$, we see that transcendental measure set can have size $\frak{c}$. By taking subsets of $S$, we can get transcendental measure set of any size at most continuum, and, clearly, not of any greater cardinality.

Answer to this question provides an algebraically independent set of real numbers of size continuum. Enumerate it as $\alpha_{i,r}$ where $i\in\{1,...,d\},r\in\Bbb R$. Now define set $S=\{(\alpha_{1,r},...,\alpha_{d,r}):r\in\Bbb R\}\subseteq\Bbb R^d$. Now, measure of any simplex with vertices on these points can be expressed using coordinates of the vertices and radicals. If any of these turned out to be an algebraic number, we would have a non-trivial algebraic relation between $\alpha_{i,r}$, which is not possible, as they are algebraically independent.

As $|S|=\frak{c}$, we see that transcendental measure set can have size $\frak{c}$.

Answer to this question provides an algebraically independent set of real numbers of size continuum. Enumerate it as $\alpha_{i,r}$ where $i\in\{1,...,d\},r\in\Bbb R$. Now define set $S=\{(\alpha_{1,r},...,\alpha_{d,r}):r\in\Bbb R\}\subseteq\Bbb R^d$. Now, measure of any simplex with vertices on these points can be expressed using coordinates of the vertices and radicals. If any of these turned out to be an algebraic number, we would have a non-trivial algebraic relation between $\alpha_{i,r}$, which is not possible, as they are algebraically independent.

As $|S|=\frak{c}$, we see that transcendental measure set can have size $\frak{c}$. By taking subsets of $S$, we can get transcendental measure set of any size at most continuum, and, clearly, not of any greater cardinality.