I would like to prove the following result :
A semi-algebraic set in $\mathbb{R}^d$ has dimension $d$ if and only if its interior is non empty
The dimension here has to be understood in the semi algebraic sense as defined now :
A semi-algebraic set $A$ in $\mathbb{R}^d$ can always be decomposed as $A=\cup_{k=1}^{n}A_k$ where the $A_k$’s are disjoint semi algebraic sets such that each of them is semi algebraically homeomorphic to $(0,1)^{d_k}$. The dimension of $A$ is then defined to be $\dim(A)=\max_{1\leq k\leq p}\{d_k\}$.
Here is my attempt :
Let $A\subset\mathbb{R}^d$ be semi-algebraic.
Suppose it’s dimension is $d$ but it’s interior is empty.
We now by the decomposition of semi algebraic set in $\mathbb{R}^d$ that
$$A =\cup_{k=1}^{p}A_k$$
Where the $A_k$’s are disjoint semi algebraic sets such that each of them is semi algebraically homeomorphic to $(0,1)^{d_k}$.
Moreover $\dim(A)=\max_{1\leq k\leq p}\{d_k\}=d$ by assumption.
Take the $A_k$ that is semi algebraically isomorphic to $(0,1)^d$. Observe that $(\frac{1}{2}, \frac{3}{4})^d$ is a proper subset of $(0,1)^d$. Now take the semi algebraic homeomorphism $h:A_k\to (0,1)^d$.
We know that $h^{-1}((\frac{1}{2}, \frac{3}{4})^d)$ is open in $A_k$, this means it is of the form
$$h^{-1}((\frac{1}{2}, \frac{3}{4})^d) = U\cap A_k$$
But $A$ has empty interior, this implies that $U\cap A_k$ is either empty or the set $A_k$, both contradicting the fact that $h$ is a bijection. We conclude that $A_k$ is not homeomorphic to $(0,1)^d$. We can repeat this argument to all the $A_k$’s homeomorphic to $(0,1)^d$, which contradicts the fact that $\dim A=d$.
For the reverse implication, let’s try by contradiction. Assume that $\mathring{A}\neq\emptyset$ but $\dim(A)\neq d$. First $\dim(A)<d$. Moreover $\mathring{A}$ is a nonempty open semi-algebraic set so $\dim(\mathring{A})=d$ by a well know property on non empty open semi-algebraic set of $\mathbb{R}^d$.Since $\mathring{A}\subset A$ we have $\dim(A)\geq\dim(\mathring{A})=d$ which is a contradiction.
For the reverse implication, the fact that $\dim(A)<d$ seems to be a shortcut, I should justify it properly I think !
If you have some remarks/hints do not hesitate please
Thank you a lot
About the fact that $\dim(A)<d$ :
We can justify this by the fact that $A\subset\mathbb{R}^d\implies\dim(A)\leq\dim(\mathbb{R}^d)=d$