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Let $M$ and $N$ be differentiable manifolds in $\mathbb{R}^{n}$, and let $p \in \mathbb{R}^{n}$. We say that $M$ and $N$ are transversal at $p$ if $$T_{p} M + T_{p}N = \mathbb{R}^{n}.$$ By dimension formula, if $M$ and $N$ are transversal at $p$, then $$n = \mbox{dim}T_{p}M + \mbox{dim}T_{p}N - \mbox{dim}(T_{p}M \cap T_{p}N).$$

We also know that transversality persists under sufficiently small perturbations.

Consider this example. Let $n=2$, $M$ be the $x$ axis, $N$ be the graph of function $f(x)=x^{3}$, and take $p=(0,0)$, i.e. the origin. $M$ and $N$ intersect at $p$ but NOT transversally because their tangent spaces at $p$ coincide and hence the dimension formula above is not satisfied.

Now let us perturb $N$ slightly. Let $\epsilon>0$ be small and let $N_{\epsilon}$ be graph of $f_{\epsilon}(x) = x^{3} + \epsilon$. Then $M$ and $N_{\epsilon}$ are now transverse at point $p_{\epsilon} = (-\epsilon^{1/3},0)$, i.e. near the original point $p$.

Conversely, if we start with $N_{\epsilon}$, $M$ and $p_{\epsilon}$ we expect transversality to persist for small $\epsilon$, but again it fails for $N$.

So what went wrong in the above example? One thing I can think of is that the perturbation is of order $O(\epsilon)$ while the new intersection is $O(\epsilon^{1/3})$ away from the original point, hence not "close" enough...

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    $\begingroup$ If you start with $N_\epsilon$, "persistence for small $\delta$" will not include $N$: $\delta$ will be less than $\epsilon$. $\endgroup$ – user98602 May 22 '16 at 15:18
  • $\begingroup$ Oh ok, how could establish an upper bound for $\delta$?, (I mean both for the specific case and in general) @MikeMiller $\endgroup$ – Alex May 22 '16 at 15:21
  • $\begingroup$ Oh ok, how could establish an upper bound for δδ?, (I mean both for the specific case and in general) @MikeMiller $\endgroup$ – Alex May 22 '16 at 18:43
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If an intersection of $M$ and $N$ is non-transversal, that means that you cannot be sure that the intersection persists under small perturbation. In your example, it's clear that the intersection will not disappear. However, if you would have chosen $N$ to be the graph of $g_\epsilon(x) = x^2 + \epsilon$, then it's clear that the intersection at $x=0$ will disappear for $\epsilon > 0$.

In the end, it's a matter of logic: $P$ (an intersection is transversal) $\to$ $Q$ (this intersection persists under small perturbations) is equivalent to the contrapositive $\neg Q \to \neg P$, but not equivalent to the inversion $\neg P \to \neg Q$.

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