set of immersions on a manifold is an open set in the set of all mappings to $\mathbb{R}^{2m+1}$

I'm studying a proof of Takens' Delay Embedding Theorem. A key fact used is that the set of immersions is open in the set of all mappings (mappings from an $m$-dimensional manifold M to $\mathbb{R}^{2m+1}$).

I don't know how dumb this question is, but in what sense is it open? In which topology do we have this openness of our set of interest? I'm trying to develop an intuition on the set of such mappings, so I ask this question.

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You'll find these details in both Guillemin and Pollack's book, and Hirsch's book on differential topology. The topology you put on this mapping space demands that maps are "close" when they are both uniformly close but also their derivatives have to be uniformly close. Once you know the derivatives are uniformly close, it's as Sammy describes. – Ryan Budney Mar 19 '13 at 5:08

It's a good question. Usually for mapping spaces, one uses the compact-open topology. It has as a subbase consisting of sets $C(K, U) = \left\{f: X \to Y \;|\; f(K) \subseteq U \right\}$, where $K$ varies over compact $K \subseteq X$ and open $U \subseteq Y$. This topology enjoys many natural properties (especially when the spaces are Hausdorff) that are itemized on the wiki page.
Exactly what constitutes a map $f:X \to Y$ depends on the category that you're interested in. It's likely that you're considering smooth maps, given the (differential-topology) tag.
Let me see if I understand your answer, the set of immersions is open in the set of smooth mappings from m-dimensional manifold $M$ to $\mathbb{R}^{2m+1}$ endowed with the compact open topology. Is there an intuitive way to explain why it is open? – a a Mar 19 '13 at 4:50