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Exercise 1291Topology › Topology of Mn(ℝ)L1Added 4 Nov 2026

Separability of $M_n(\mathbb{R})$

#separability#dense countable subset#rational matrices

Statement

Show that Mn(R)M_n(\mathbb{R}) is separable.

Hint

Consider matrices with rational entries.

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Solution

Countable dense subset. Let Mn(Q)M_n(\mathbb{Q}) be the set of matrices with rational entries. It is countable (finite product of countable sets).

Density. For any A∈Mn(R)A\in M_n(\mathbb{R}) and ε>0\varepsilon>0, approximate each entry aija_{ij} by a rational qijq_{ij} with ∣aij−qij∣<ε|a_{ij}-q_{ij}|<\varepsilon. The matrix Q=(qij)Q=(q_{ij}) satisfies ∥A−Q∥∞<ε\|A-Q\|_\infty<\varepsilon.

Conclusion. Mn(Q)M_n(\mathbb{Q}) is countable and dense in Mn(R)M_n(\mathbb{R}), which is separable.