Show that the function $x \mapsto \dim(\mathrm{Im}(D_{x}f))$ is locally increasing and upper semicontinuous

Question

Let $f:\mathbb{R}^{m} \to \mathbb{R}^{n}$ of class $C^1$. Show that the function $x \mapsto \dim(\mathrm{Im}(D_{x}f))$ is locally increasing.

Lemma. Let $M$ a matrix with rank $r$. Then there is $\delta>0$ such that $\Vert M - N \Vert < \delta$ implies $\mathrm{rank}(N) \geq r$.

Proof. Obviously the result follows for $r = 0$. If $r > 0$, there is a $r \times r$ invertible submatrix $T$ in $M$. Since $T$ is invertible, $\det(T) \neq 0$. The determinant is a continuous function, so there is $\delta > 0$ such that for every $r \times r$ matrix $V$ with $\Vert T - V \Vert < \delta$ we have $\det(V) \neq 0$.

Here is my first question: If $N$ is a matrix with a submatrix $V$ that satisfies the previous description, so $\Vert T - V \Vert \leq \Vert M - N \Vert < \delta$ whenever $\Vert M - N \Vert < \delta$, that is, $\det(V) \neq 0$ then $\mathrm{rank}(N) \geq r$. But how can I ensure that a matrix $N$ with $\Vert M - N \Vert < \delta$ has such submatrix?

I think that this lemma solves the question, I dont write so, correct me if I'm wrong.

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Show that $x \mapsto \dim(Im(D_{x}f))$ is upper semicontinuous.

I dont know how to use the previous question to prove it. I have the definition:

A function is upper semicontinuous in $x_{0}$ if for each $\epsilon > 0$, there is a $\delta > 0$ such that $f(x) < f(x_{0}) + \epsilon$ for any $x \in B_{\delta}(x_{0})$.

Answer 1

Concerning your first question. You didn’t specify the norm, but I guess that Lemma holds for any common norm on real-valued matrices, for instance, for $\ell^\infty$-norm, which concerns my following remarks. Your proof of Lemma is correct and such a submatrix $T$ exists (see, for instance, subsection “Determinantal rank – size of largest non-vanishing minor” here). Then also exists respective $\delta>0$. Now if $\Vert M - N \Vert < \delta$ then a submatrix $V$ of $N$ with the same indices of entries which has the matrix $T$ in $M$ (so $V$ exists because $T$ exists) satisfies $\Vert T - V \Vert \leq \Vert M - N \Vert < \delta$. Then $\det(V) \neq 0$ and $\mathrm{rank}(N) \geq r$.