In Game Theory by Fudenberg and Tirole, they define the set of rationalizable strategies in the following way.
Definition 2.3 in FT: Set $\tilde{\Sigma}_{i}^{0}=\Sigma_{i}$ where $\Sigma_{i}$ is the space of mixed strategies for player $i$. For each $i$, define recursively $$\tilde{\Sigma}_{i}^{n}=\{\sigma_{i}\in \tilde{\Sigma}_{i}^{n-1}|\exists \sigma_{-i} \in \times_{j\not=i} \text{convex hull} \,\left(\tilde{\Sigma}_{j}^{n-1}\right) s.t \,\,u_{i}(\sigma_{i},\sigma_{-i})\geq u_{i}(\sigma_{i}',\sigma_{-i}) \,\,\forall \sigma_{i}'\in \tilde{\Sigma}_{i}^{n-1}\}$$ The rationalizable strategies for player $i$ are $R_{i}=\cap_{n=0}^{\infty}\tilde{\Sigma}_{i}^{n}$
The convex hull of a set $X$ is the smallest convex set that contains $X$
My question is that, why do we need the convex hull operator in the definition?
I think $\tilde{\Sigma}_{i}^{n}$ is the set of strategies that survive the $n-1$ rounds of deletion of never-best-response strategies, I don't know why we need convexity here.