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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 $A$?

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The answer is 'yes'.

  1. Let $ I $ be a right ideal of $ B $.

  2. Let $ a \in A $.

  3. 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 $.

  4. Let $ y \in {\text{cl}_{A}}(I) $.

  5. There exists a directed set $ I $ and a net $ (y_{i})_{i \in I} $ in $ I $ that converges to $ y $.

  6. Observe that $ I \times \Lambda $ is a directed set equipped with the product partial ordering.

  7. 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.

  8. 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) $.

  9. 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.