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Is the set of nilpotent $m \times m$ real matrices compact?

I found the proof of this statement, using Heine-Borel theorem on $\mathbb R^n$. Tha'ts quite good.

But, is it possible to prove this statement via the following definition of compactness?

A set is said to be compact if every open cover admits a finite subcover.

Assume the usual topology.

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  • $\begingroup$ mmm.. but via opensets how could i start the proof?? $\endgroup$ – David Mar 4 '15 at 3:21
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    $\begingroup$ It is quite surprising that you found a proof of this fact, as it is false! $\endgroup$ – Mariano Suárez-Álvarez Mar 4 '15 at 3:21
  • $\begingroup$ Yes, well that would be a problem. Yes, this is false. For example, $\begin{pmatrix} 0 & n\\ 0 & 0\end{pmatrix}$ is nilpotent for every $n$, so the set of nilpotent matrices is not bounded. $\endgroup$ – Moya Mar 4 '15 at 3:25
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    $\begingroup$ Heine Borel says a set is compact in Euclidean space if and only if it is closed and bounded. $\endgroup$ – Moya Mar 4 '15 at 3:29
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    $\begingroup$ You can exhibit an open covering that has no finite subcovering. Think about trying to find open balls containing matrices of the form I wrote above (in $\mathbb{R}^4$) that are small enough such that they are disjoint. Then you can't refine this cover. $\endgroup$ – Moya Mar 4 '15 at 3:36
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The set $N$ of $m \times m$ real nilpotent matrices is compact iff $m = 1$, in which case $N$ is the singleton containing the zero matrix (in particular it is finite, hence compact).

For $m > 1$, consider the norm covering of $M(m, \mathbb{R})$ by open balls, namely the covering $\mathcal{B} := \{B_r : r \in \mathbb{Z}_+\}$, where $$B_r := \{A \in M(m, \mathbb{R} : ||A|| < r \},$$ and $||A||$ is (for example) $\sqrt{\sum_{i, j = 1}^m A_{ij}^2}$.

Now, by construction, $\{B_r \cap N : i \in \mathbb{Z}_+\}$ is an open cover of $N$, and we will show it admits no finite subcover: The sets $B_r$ are nested, that is, $B_1 \subseteq B_2 \subseteq B_3 \subseteq \cdots$, and hence so are the sets $B_r \cap N$; thus, the union of the sets in any finite subcollection $\{B_{r_1} \cap N, \ldots, B_{r_k} \cap N\} \subset \mathcal{B}$ is just $B_R$, where $R := \max\{r_1, \ldots, r_k\}$. On the other hand, the matrix $R_0$ with $(1, 2)$ entry $R$ and all other entries zero is in $N$ (its square is the zero matrix), but $||R_0|| = R$, so $R_0$ is not in $B_R$, so hence arbitrary finite subcollection is not a cover of $N$, that is, $N$ is not compact.

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