We say (according to <https://ncatlab.org/nlab/show/%28infinity%2Cn%29-category+with+duals>) that a symmetric monoidal $(\infty,1)$ category $\mathcal{C}$ *has duals* if its homotopy category $h\mathcal{C}$ is rigid monoidal. 


I'm interested in the $\infty$-category $Sp$ of spectra. What is the largest stable $\infty$-subcategory $\mathcal{C}$ of $Sp$ such that $\mathcal{C}$ has duals? Is such a category nontrivial? If so does it have a nice description? In this case the dual object of $X \in Sp$ is $Map(X, \mathbb{S})$, where $\mathbb{S}$ is the sphere spectrum $\Sigma^{\infty} S^0$. 



Dual objects in the $\infty$-category of spectra
As requested, the comments turned into answers:


1. The dualizable objects in spectra are precisely the finite spectra (i.e. spectra of the form $\Sigma^{-k}\Sigma^{\infty}X$ where $X$ is a finite complex.)
2. If you only want the statement 'dualizable objects are finite spectra and their retracts' there is a very formal proof that works in great generality: (a) the unit in $\mathsf{Sp}$ is a compact and $\mathsf{Sp}$ is closed symmetric monoidal, (b) every object in $\mathsf{Sp}$ is a filtered colimit of finite spectra; it follows that every dualizable object $X$ is compact and that the identity map $X \to X$ factors through a finite spectrum, hence $X$ is a retract of a finite spectrum.
3. If you want to know that retracts of finite spectra are themselves finite, this is a bit less formal. One may argue directly or use (2) to show that, for $X$ dualizable, $H\_\*(X, \mathbb{Z})$ is finitely generated and projective, hence free and concentrated in finitely many degrees. Now inductively build cell complexes $Y\_j$ and maps $Y\_j \to X$ which are equivalences on $H\_m(-,\mathbb{Z})$ for $m\le j$. This stops at some finite step and you get a map $Y \to X$ from a finite complex to $X$ which is an equivalence on $H\_\*(-,\mathbb{Z})$. But $X$ is connective (lots of ways to see this... for example, by (2) it's a retract of something connective), so this map is an equivalence by Hurewicz. (Note: for spaces, this argument works as soon as $X$ is simply connected, but can fail because of Wall's finiteness obstruction when $X$ is not simply connected).


Obligatory: none of this had anything to do with $\infty$-categories.


