$\newcommand{\GLp}{\operatorname{GL}_n^+}$
$\newcommand{\SO}{\operatorname{SO}_n}$
Consider the set
$$ X=\{ A \in \GLp \,\,|\,\, \text{ all the singular values of } A \text{ are distinct } \},$$
where $\GLp$ is the group of real $n\times n$ invertible matrices with positive determinant.
Is $X$ connected? (with the subspace topology induced by $\GLp$). Is $X$ dense in $\GLp$?
Since the singular values of $A$ are the eigenvalues of $\sqrt{A^TA}$, I guess this might be connected to the question:
Are matrices in $\GLp$ with distinct eigenvalues dense (in $\GLp$) and connected? I am quite sure they should be dense, but I am not sure about the connectedness.
Edit:
I am quite sure that the density is OK. Indeed, the singular values $\sigma_i$ are distinct if and only if their squares $\sigma_i^2$ are distinct. The $\sigma_i^2$ are the eigenvalues of $A^TA$. Now the argument here with the discriminant of the characteristic polynomial of $A^TA$ works.
Here is an idea regarding the connectedness:
Let $A=U\Sigma V^T \in \GLp$ be with distinct singular values. Since we can always assume that $U,V \in \SO$ and $\SO$ is connected, we can always connect $A$ to $\Sigma$ by "moving on the sides" in $\SO$. (Since we are only changing the orthogonal components, the singular values remain constant along this path).
So, we only need to move between the different $\Sigma$'s. That is, we are left with the following question:
Is the set of vectors in $(\mathbb{R}^{\ge 0})^n$ with different entries connected? I guess there should a be a slick argument deciding this either way.
Best Answer
To answer your second question: no, the subset of $(\mathbb R^{\ge0})^n$ containing vector with distinct coordinates is not necessarily connected. Consider the case where $n=2$. The first quadrant of the $x-y$ plane is divided by the line $x=y$ into two connected components.
But the answer to your first question is yes, because we can perform singular value decomposition on every $A\in X$ so that the singular values are arranged in decreasing order (and the orthogonal matrices have determinants $1$). Since any convex combination of two strictly decreasing tuples is again a strictly decreasing tuple (and, as you said, $SO_n$ is path-connected), $X$ is path-connected. As each decreasing tuple is the limit of a sequence of strictly decreasing tuples, $X$ is dense in $GL_n^+$.