# Why does this imply $F(k,n)$ is open?

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 $$\mathbb{R}^{k\cdot n}$$?

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

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.