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Let $S \subset M_n(\mathbb R)$ be an open subset and $0 \notin S$. Suppose if $A \in S$, we can construct a continuous path within $S$ between $A$ and $A(\alpha) \in S$, where $\alpha \in \mathbb R_+$ is a parameter (can be freely chosen) that scales all entries of $A$ but in different exponents. That is, some entries are scaled at $\alpha$, some are at $\alpha^2, \alpha^3, \dots, \alpha^n$ etc but at least $\alpha$. So we can make the norm of $A(\alpha)$ to be arbitrarily small by choosing $\alpha$ sufficiently small.

Now I would like to prove $S$ is connected. Here is my argument: Let $A_1, A_2 \in S$. Let $a = \max_{i, j}((A_1)_{ij}, (A_2)_{ij})$. For sufficiently small $\varepsilon > 0$, we choose $c = \varepsilon/(n^2 a)$, then we know $\|A_1(c)\|_F < \varepsilon, \|A_2(c)\|_F \le \varepsilon$. So $\|A_1(c)-A_2(c)\|_F < 2\varepsilon$. Ideally I want to conclude $A_1(c), A_2(c)$ are connected by openness of $S$. This way we have a continuous path \begin{align*} A_1 \xrightarrow{} A_1(c) \xrightarrow{} A_2(c) \xrightarrow{} A_2. \end{align*} But this does not feel right. Since if we want to use openness, we fix an element $B$ first and there exists some $\varepsilon$-ball of $B$. But here the $\varepsilon$ is chosen first. Then we get $A(\varepsilon)$ and the openness only guarantees another $\varepsilon'$. But since we can make the norm to be arbitrarily small, is it possible to fix this argument?

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  • $\begingroup$ Do I understand correctly that you want to prove that each open $S$ not containing $0$ is connected? That would obviously be wrong. $\endgroup$ – Paul Frost Aug 7 '18 at 8:02

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