Let $A$ be a topological unital algebra and let $B$ be its dense subalgebra with unit. Let $I$ be a right ideal of $B$. Is the closure of $I$ a right ideal of $A$?
Dense subalgebras of topological algebras
1 Answers
The answer is 'yes'.
Let $ I $ be a right ideal of $ B $.
Let $ a \in A $.
By the denseness of $ B $ in $ A $, there exists a directed set $ \Lambda $ and a net $ (b_{\lambda})_{\lambda \in \Lambda} $ in $ A $ that converges to $ a $.
Let $ y \in {\text{cl}_{A}}(I) $.
There exists a directed set $ I $ and a net $ (y_{i})_{i \in I} $ in $ I $ that converges to $ y $.
Observe that $ I \times \Lambda $ is a directed set equipped with the product partial ordering.
Consider the net $ (y_{i} b_{\lambda})_{(i,\lambda) \in I \times \Lambda} $, which is made up of elements of $ I $ because $ I $ is a right ideal.
As algebra multiplication is continuous, we see that $ \displaystyle ya = \lim_{(i,\lambda) \in I \times \Lambda} y_{i} b_{\lambda} \in {\text{cl}_{A}}(I) $.
Therefore, $ {\text{cl}_{A}}(I) $ is a right ideal of $ A $, by the arbitrariness of $ y $ and $ a $.
It is also necessary to check that $ {\text{cl}_{A}}(I) $ is a linear subspace of $ A $, but this is easy to show, using an argument with nets again.