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Here, we have $F(k,n)$ defined as the set of ordered $k$-tuples of linearly independent vectors in $\mathbb{R^n}$.

To start, let $X \in F(k,n)$. We can express $X$ as

$$X = \begin{bmatrix} x_1^1 & \cdots & x_1^n \\ \vdots & \vdots & \vdots \\ x_k^1 & \cdots & x_k^n \end{bmatrix}$$

where this is a matrix of rank $k$. Having rank $k$ implies that not all $k \times k$ minors are equal to zero. Define a function:

$$f(X) = \sum(k\times k \text{ minors of } X)^2$$ This is a continuous function from $\mathbb{R}^{k\cdot n} \to \mathbb{R}$.

Using definition of continuity, we get that $f^{-1}(\{0\})$ is a closed subset of $\mathbb{R}^{n\cdot k}$.

My question is, why does $f^{-1}(\{0\})$ being closed imply that $F(k,n)$ is an open subset of $\mathbb{R}^{k\cdot n}$?

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  • $\begingroup$ Yes. The complement of a closed set is open. $\endgroup$ – N. S. Jan 26 at 20:39
  • $\begingroup$ Yes, I know this, but I didn't make the connection that $F(k,n)$ is the complement of $f^{-1}(\{0\})$. Thanks. $\endgroup$ – Sorey Jan 26 at 20:40
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Yes. The complement of a closed set is open. Alternately, $$F(k,n)= f^{-1}( (- \infty,0) \cup (0, \infty)$$ is the pre-image of an open set under a continuous function.

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