I have a conjecture like this $$\lim_{n\rightarrow\infty}\min_{m\in\{1,2,3\}}a_n(m)=\min_{m\in\{1,2,3\}}a(m)$$ where $a(m)=\lim_{n\rightarrow\infty}a_n(m)$ for any $m\in\{1,2,3\}$ and $n$ is interger and $a_n(m)$ is a real number function indexed by $m$ and $n$.
My proof step is as follows: Since $a_n(m)$ is a point-wise convergent function, we have $\forall\,\epsilon_1,\epsilon_2,\epsilon_3>0$, there exists $N_1,N_2,N_3>0$ such that for all $n>\max\{N_1,N_2,N_3\}$, $|a_n(m)-a(m)|\leq\epsilon_m\leq\max_{m\in\{1,2,3\}}\epsilon_m=\epsilon'$ for any $m\in\{1,2,3\}$.
Therefore, for any $n>\max\{N_1,N_2,N_3\}=N'$, we have $$\min_{m\in{1,2,3}}a(m)-\epsilon'=\min_{m\in{1,2,3}}(a(m)-\epsilon')\leq\min_{m\in\{1,2,3\}}a_n(m)\leq\min_{m\in{1,2,3}}(a(m)+\epsilon')=\min_{m\in{1,2,3}}a(m)+\epsilon'$$ That is, we have $|\min_{m\in\{1,2,3\}}a_n(m)-\min_{m\in\{1,2,3\}}a(m)|\leq\epsilon'$ for all $n>N'$. Consequently, since $\epsilon'$ and $N'$ can be any chosen numbers, we have $$\lim_{n\rightarrow\infty}\min_{m\in\{1,2,3\}}a_n(m)=\min_{m\in\{1,2,3\}}a(m)$$
Is this proof correct?
Edit: If the equality is wrong, what should be verified? "$\leq$" or "$\geq$?"
The equality is right. You can also use the fact that for finite $k$ the function, $$(x_1, \ldots, x_k) \mapsto \min_{j=1,\ldots,k} x_j$$ is continuons.