Dropping closed condition in Transversality Theorem

Most theorems stating that transversality is an open condition look like:

We have manifolds $$M$$, $$N$$, and a submanifold $$A\subseteq N$$. Then assuming $$A$$ is closed in $$N$$, in the appropriate topology on the space of smooth maps $$C^\infty(M,N)$$ (weak/compact-open topology if $$M$$ is compact, or strong/Whitney topology if $$M$$ is noncompact), the maps transverse to $$A$$ form an open dense subset.

However, I don't understand why this condition $$A$$ closed in $$N$$ is needed. I was trying to see if there are counterexamples for it.

For reference, here is Hirsch's statement (Differential Topology, Chapter 3):

And also from this bachelor thesis https://www.math.harvard.edu/media/ThesisXFinal.pdf:

Obs: This post (Reference for transversality -- relative version) seems related, but I'm not sure it answers this question. The answer in it seems to explain why $$A$$ closed in $$N$$ matters, but only if we're not asking transversality in all of $$M$$.

Let $$M=\mathbb{R}$$, $$N=\mathbb{R}^2$$, and $$A \subset N$$ be the 1-submanifold consisting of the $$x$$-axis and a sequence of open vertical lines segments approaching the $$y$$-axis. Consider the element of $$C^\infty(M,N)$$ which just embeds $$M$$ as the $$y$$-axis. This map is transverse to $$A$$.
However, any open set in $$C^\infty(M,N)$$ around this map will include embeddings of $$M$$ that intersect one of the vertical line segments tangentially. So the transverse maps do not form an open subset of $$C^\infty(M,N)$$. Note thought that they are still dense, as any of the embeddings of $$M$$ that shares a vertical tangency with $$A$$ can be nudged a bit to become transverse again.
Easier example: Take $$A$$ to be the positive real axis in $$N=\Bbb R^2$$. The function $$f\colon \Bbb R\to \Bbb R^2$$ given by $$f(t)=(t,t^2)$$ is transverse to $$A$$, but no neighborhood of $$f$$ in any topology will stay transverse to $$A$$. (You can easily modify this with maps $$S^1\to \Bbb R^2$$, if you want a compact domain.)
• I think you mean transverse to $A$? Also the parabola is not transverse to the positive real axis, right? Commented Nov 6, 2021 at 17:43
• Positive means positive, not non-negative. So, yes, it's transverse by default. Yes, I meant $A$. I'll edit that. Commented Nov 6, 2021 at 17:54