I am trying to go through the classical paper by Nash on the existance of $C^1$ isometric immersion of a Riemannian manifold $(M,g)$ (here is the Jstor link: https://www.jstor.org/stable/1969840?seq=1#metadata_info_tab_contents). Let me recall the setup so I can make my question as accurate as possible:

We have a (possibly non-compact) Riemannian manifold $(M,g)$ and a short immersion $z:M\rightarrow \mathbb{R}^k$ wich induces a new metric $h$ on $M$. Shortness means that the tensor $\delta=g-h$ is again possitive definite.

We also have an open cover $N_p$ along with a subordinate partition of unity $\phi_p$ such that every neighborhood $N_p$ meets only finite others. So we now focus on the immersion of $N_p$ into $\mathbb{R}^k$.

Now Nash wants to approximate the tensor $\frac{1}{2}\delta_{ij}$ by another positive tensor $\beta_{ij}$. He does this by first finding a set $M_{\mu,\nu}$ of p.d.s.m (possitive definite symmetric matrices) such that for any p.d.s.m. $A$ we have $A=\sum_{\mu,\nu}C^{*}_{\mu,\nu}M_{\mu,\nu}$ where the coefficients $C^*_{\mu,\nu}$ depend smoothly on $A$.

Here is what is bugging me. Nash writes:

So if I understand correctly, since $\beta_{ij}$ is a smooth choice of a p.d.s.m for any point $x\in N_p$, we can view $\beta_{ij}$ as smooth immersion of $N_p$ into the space of p.d.s.m's , which is an open cone in the vector space of symmetric matrices. My question is: **Why can't we choose $\beta_{ij}=\delta_{ij}$?**

Again, on top of page $391$ it is stated that we can make $\beta_{ij}$ arbitrarily close to $\delta_{ij}$. But what stops us from having them exactly equal? Any expository notes on this theorem will also be greatly appreciated!