**Theorem 1.**Let G be a group which has a complete metric topology for which multiplication is jointly continuous. Then inversion is continuous (i.e., G is a topological group).

There is a short, slick proof in Pfister, "Continuity of the inverse", Proc AMS 95(1985) 312-314, though the result is older.

For the rest of this note I follow Schweitzer, Larry B. “A Short Proof that $M_{n}(A)$ is local if $A$ is local and Fr\'echet.” International Journal of Mathematics 3 (1992): 581-589. Suppose that B is a C* or Banach algebra and that A is a dense subalgebra which is a Fr\'echet algebra under some topology stronger than the topology of B.

**Theorem 2.**A is holomorphically closed in B iff it is inverse closed.

For the proof, one notes that inverse closure implies that the invertibles are open in A, so their topology (in A) can be given by a complete metric. By Theorem 1, inversion is continuous. This means that the Cauchy integral formula for the holomorphic calculus converges in A.

**Theorem 3.**A is inverse closed in B iff every irreducible A module is a submodule of some B module.

This involves some topological-algebraic manipulation with maximal left ideals. An irreducible A module is of the form $A/m$, where $m$ is some maximal left ideal. Let M be the closure of m in B. Using density and inverse closure one sees that $m=M\cap A$. But then

$A/m = A/A\cap M = (A+M)/M \subseteq B/M $ is contained in the B-module $B/M$.

**Corollary.**$A$ is inverse (or holomorphically) closed in $B$ iff every matrix algebra $M_n(A)$ is inverse closed in $M_n(B)$.

This follows since an irreducible $M_n(A)$-module is just the direct sum of $n$ copies of some irreducible $A$-module.

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