## Appendix A Proofs of Lemmas 1 and 2
Furthermore, the sequence \((x_{m}^{\prime})_{m\in\mathbb{N}}:=(\mathcal{S}_{\infty}f_{m}^{\prime})_{m\in \mathbb{N}}\) is a sequence of elementary computable sequences. In the following, we will prove by case distinction that for all \(m\in\mathbb{N}\), the sequence \((x_{m,k})_{k\in\mathbb{N}}\) converges effectively towards \(x_{m}^{\prime}\) in \(\ell_{0}^{\infty}\), with respect to the \(\mu\)-recursive modulus of convergence \(\xi:\mathbb{N}^{2}\to\mathbb{N}\), \((m,K)\mapsto K\). First, assume that \(m\in\Omega\) is satisfied. Then, there exists \(k\in\mathbb{N}\) such that for all \(K\in\mathbb{N}\) that satisfy \(K\geq k\), we have \(x_{m,K}=x_{m,k}\). Consider the smallest such \(k\in\mathbb{N}\) and observe the following for all \(K\in\mathbb{N}\):  
* If \(K\in\mathbb{N}\) satisfies \(K\geq k\), we have \(\|x_{m}^{\prime}-x_{m,K}\|_{\infty}=\|x_{m,k}-x_{m,k}\|_{\infty}=0<2^{-K}\). * If \(K\in\mathbb{N}\) satisfies \(K<k\), we have, by the properties of the runtime function \(g_{\Omega}\) and the construction of \(h\) as above, \(h(m,k)\geq h(m,K)=L(K)+1\) with \(L(K)\) as in (A3). Accordingly, by \(x_{m}^{\prime}=x_{m,k}\) and application of the triangle inequality, we also have \[\left\|x_{m}^{\prime}-x_{m,K}\right\|_{\infty}\leq\left\|S_{\infty} f_{h(m,K)}\right\|_{\infty}+\left\|S_{\infty}f_{h(m,K)}\right\|_{\infty}\;\cdots\] \[\qquad\cdots\leq\frac{1}{h(m,k)}+\frac{1}{L(K)+1}\leq\frac{1}{2^{K +2}}+\frac{1^{K+2}}{2^{K+2}}=\frac{2}{2^{K+2}}=\frac{1}{2^{K+1}}<2^{-K}.\]Second, assume that \(m\in\mathbb{N}\setminus\Omega\) is satisfied. Then, we have \(\mathbf{x}^{\prime}_{m}\equiv 0\). Observe the following for all \(K\in\mathbb{N}\):  
* With \(L(K)\) as in (A3), we have \(\|\mathbf{x}^{\prime}_{m}-\mathbf{x}_{m,K}\|_{\infty}=\|\mathbf{x}_{m,K}\|_{\infty}\leq(L( K)+1)^{-1}<2^{-K}\). Following the preceding case distinction, we conclude that \(\|\mathbf{x}^{\prime}_{m}-\mathbf{x}_{m,K}\|_{\infty}<2^{-K}\) is satisfied for all \(m,K\in\mathbb{N}\). In other words, for all \(m\in\mathbb{N}\), \((\mathbf{x}_{m,K})_{k\in\mathbb{N}}\) converges effectively towards \(\mathbf{x}^{\prime}_{m}\) in \(\ell_{0}^{\infty}\), with respect to the \(\mu\)-recursive modulus of convergence \(\delta:\mathbb{N}^{2}\to\mathbb{N},(m,K)\mapsto K\). Consequently, \((\mathbf{x}^{\prime}_{m})_{m\in\mathbb{N}}\) is a computable sequence of sequences in \(\mathcal{C}\ell_{0}^{\infty}\). It remains to show that for all \(m\in\Omega\), we have \(\|f^{\prime}_{m}\|_{\infty}\geq 1\), while for all \(m\in\mathbb{N}\setminus\Omega\), we have \(\|f^{\prime}_{m}\|_{\infty}=0\), which we again prove by case distinction. First, recall that if \(m\in\Omega\) is satisfied, \(f^{\prime}_{m}\) is an elementary computable function such that for some \(M\in\mathbb{N}\), we have \(f_{M}=f^{\prime}_{m}\). By assumption, \(C\geq\|f_{M}\|_{\infty}\geq f_{M}(\
icefrac{{1}}{{2}})\geq 1\) is satisfied for some number \(C>0\) and all \(M\in\mathbb{N}\). Hence, for all \(m\in\Omega\) there exists an \(M\in\mathbb{N}\) such that we have \(C\geq\|f^{\prime}_{m}\|_{\infty}=\|f_{M}\|_{\infty}\geq 1\). Second, recall that if \(m\in\mathbb{N}\setminus\Omega\) is satisfied, then \(f^{\prime}_{m}\) is the trivial elementary computable function. That is, we have \(f^{\prime}_{m}\equiv 0\), in which case \(\|f^{\prime}_{m}\|_{\infty}=0\) holds true. Hence, for all \(m\in\mathbb{N}\setminus\Omega\), we have \(\|f^{\prime}_{m}\|_{\infty}=0\). Following the preceding case distinction, we conclude that for all \(m\in\Omega\), we have \(\|f^{\prime}_{m}\|_{\infty}\geq 1\), while for all \(m\in\mathbb{N}\setminus\Omega\), we have \(\|f^{\prime}_{m}\|_{\infty}=0\).