If for each $i=1,\ldots,I$, $X_{i}^{n}=(x_{i1}^{n},\ldots x_{iJ}^{n})$ is a bounded sequence such that $\left|\left|X_{i}^{n}\right|\right|<M$ for some finite $M$, can we use the Bolzano-Weierstrass theorem to conclude that there exists a subsequence $n_{k}$ such that the whole $I\times J$ matrix $X^{n_k}$ converges to some other $I\times J$ matrix $X$?
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$\begingroup$ Yes, it is still a finite dimensional vector space. $\endgroup$– copper.hatCommented Sep 20, 2016 at 2:47
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$\begingroup$ Thanks. Would the argument be applying the B-W to each row sequentially and choosing a convergent subsequence of the previous subsequence? $\endgroup$– SubmartingaleCommented Sep 20, 2016 at 2:49
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$\begingroup$ You could do it that way if you like. $\endgroup$– copper.hatCommented Sep 20, 2016 at 2:50
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$\begingroup$ I see. So this would follow directly from the theorem given that we have a finite-dimensional vector space - no need for an argument. $\endgroup$– SubmartingaleCommented Sep 20, 2016 at 2:51
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1$\begingroup$ Yes. It is easy to exhibit an isomorphism with $\mathbb{R}^N$ for $N=m \times n$ (or $\mathbb{C}^N$). The key is that there is a finite basis. The Frobenius norm and the trace inner product are useful tools in this regard. $\endgroup$– copper.hatCommented Sep 20, 2016 at 2:53
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Yes. Finite dimensional real or complex vector spaces are locally compact. In fact all closed bounded sets, in such spaces, are compact.
If $S=(y_{i,j})_{i=1,...,m; j\in \mathbb N}$ is a bounded sequence (with $y_{i,j}$ in $\mathbb R$ or in $\mathbb C$), let $(y_{1,f_1(j)})_j$ be a convergent subsequence of $(y_{1,j})_j$. Let $(f_2(j))_j$ be a subsequence of $(f_1(j))_j$ such that $(y_{2,f_2(j)})_j$ converges; if $m\geq 3$ let $(f_3(j))_j$ be a subsequence of $(f_2(j))_j$ such that $(y_{3,f_3(j)})_j$ converges, and so on, until we reach $(f_m(j))_j.$ Then $(y_{i,f_m(j)})_j$ converges for each $i\in \{1,...,m\}.$
answered Sep 20, 2016 at 3:08