I did the following differential topology exercise and I wanted to make sure everything was alright , the exercise is from Hirsch
The limit set $L(f)$ of $f:M\rightarrow N$ is the set of $y\in N$ such that $y=\lim_{n\rightarrow \infty}f(x_n)$ where $x_n$ has no convergent subsequences. Show that if dim$ N$>$2$dim$M$ and $1\leq r\leq \infty$ then embeddings are dense in $\mathcal{L}=\{f\in C_S^r(M,N): f(M)\cap L(f)=\emptyset\}$.
Now first we show that there is an open set $N_0\subset N$ such that $f(M)\subset N_0$ and $f:M\rightarrow N_0$ is a proper map. From the way the set $\mathcal{L}$ is constructed , my intuition told me that this set would be $N-L(f)$ .
This is the part I have doubts :
Indeed first we show that $L(f)$ is closed . Let $z\in \bar L(f)$ we have that there exists $y_n\in L(f)$ such that $y_n\rightarrow z$, and for each of these $y$ there exists a sequence with no convergent subsequences $x_k^n$ such that $y_n=\lim_{k\rightarrow \infty}x_k^n$. Now here I am not interely sure how to construct the sequence I was thinking first something like $x^n_n$ but I am not sure I can ensure that this has no convergent subsequences. So I have no better ideas on how to show that the set is closed.
This part I belive it's fine:
Now let's check that $f:M\rightarrow N_0 $ is in fact proper. Take a compact set $K$ in $N_0$ and suppose that $f^{-1}(K)$ is not compact so that there would existence a sequence $x_n$ with no convergent subsequences, but since $K$ is compact we would have that $f(x_{n_k})\rightarrow y$, and this would give us a contradiction so the map is proper. Now noticing that $N_0$ is a manifold of dimension greater than $2$dim$M$ , since it is an open set of $N$, we know that embeddings are dense in proper maps and so for any neighborhood of this map there exists an embedding, and so we get our desired result.
Thanks in advance !