Let $V$ be a real $d$-dimensional vector space ($d>2$). Let $1 \le r_1 <r_2 < \dots < r_k = d$ be a fixed finite sequence of numbers.
Define $H=\cup_{i=1}^kH_{r_i}$, where $H_{r}=\{ A \in \text{End}(V) \mid \operatorname{rank}(A) = r \}$.
Is $H$ an embedded submanifold of $\text{End}(V)$?
I know that each $H_{r_i}$ is a connected embedded submanifold of $ \text{End}(V)$, and that $H$ is also connected. (We can "fall in the rank" via a continuous path).
Note that I do not assume that $r_{i+1}=r_i+1$. In other words, I am allowing "jumps" of size greater than $1$ between ranks.
(The case where the ranks do jump in $1$ every time is easy; $H$ is just the space of matrices with rank bigger than or equal to $r_1$, which is an open set inside the vector space of all matrices).
No, this is false. Consider for example $r_1=1$ and $r_2=d$. Then a matrix $A\in H_d$ is invertible, and all matrices close by as well. Hence if $H_1\cup H_d$ is a submanifold, it is a submanifold of codimension zero. That is $H_1\cup H_d$ is an open subset of all matrices.
This is not the case: We can perturb the matrix $\mathrm{diag}\,(1,0,\ldots,0)\in H_1$ to lie outside $H_1\cup H_d$, namely $\mathrm{diag}\,(1,\epsilon,0,\ldots,0)\not \in H_1\cup H_d$ for all $\epsilon>0$.