If $\omega \in \beta \Bbb N\setminus \Bbb N$,we define $S_{\omega}=\{(x_n) \in \prod M_n(\Bbb C):lim_{n \to \omega}tr_n(x_n)=0\}$
Is the intersection $\cap_{\omega \in \beta \Bbb N \setminus \Bbb N}S_{\omega}$ empty?
2026-03-27 11:24:36.1774610676
intersection of sets corresponding to free ultrafiler
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No. Take $x=(x_n)$ where $x_n=\tfrac1n\,I_n$. Then $\operatorname{tr}_n(x_n)=\tfrac1n\to0$, so $x\in S_\omega$ for all $\omega$.