Let \(K\subset \mathbb{N}\) and denote by \(K(m,n)\) the cardinality of the set of elements in \(K\cap \{m,m+1,\dots,n\}\). The upper asymptotic density of \(K\) is defined by \[\overline{d}(K):= \limsup\frac{K(1,n)}{n},\] and upper uniform density of \(K\) by \[\overline{u}(K):= \limsup\frac{\max\{ K(i+1,i+n)\mid i\geq 0\}}{n}.\] Let \(x=\{x_n\}\) be a sequence of complex numbers. A number \(l\) is called a ``uniform statistical limit point'' of \(x\) if there is a sequence \(\{n_k\}\) such that \(\lim_{k\to \infty} x_{n_k}=l\) and \(\overline{u}(\{n_k\mid k\in\mathbb{N}\})>0\). Replacing upper uniform density by asymptotic density, we obtain a ``statistical limit point'' of \(x\). The paper contains the following results for a bounded sequence \(x\) of complex numbers. \begin{itemize} \item[(1)] Given a nonempty \(F_\sigma\) subset \(M\) of limit points of \(x\), there is a subsequence \(y\) of \(x\) such that \(M\) is the set of all statistical limit points of~\(y\). \item[(2)] Given a nonempty \(F_\sigma\) subset \(M\) of limit points of \(x\), there is a subsequence \(y\) of \(x\) such that \(M\) is the set of all uniform statistical limit points of~\(y\). \item[(3)] Identify the subsequences of \(x\) with the real numbers \(t\in (0,1]\) written in binary expansion with infinitely many ones and denote by \(x(t)\) the subsequence corresponding to \(t\). Then the set of \(t\in(0,1]\) for which the sets of uniform statistical limit points of \(x\) and \(x(t)\) coincide is of Lebesgue measure \(1\) or \(0\) (both cases may occur). \end{itemize}