Dense subalgebras of topological algebras 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 $A$?
 A: 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.
