Sum of Closures is Subset of Closure of Sum in Topological Vector Space

Theorem

Let $K$ be a topological field.

Let $X$ be a topological vector space over $K$.

Let $A, B \subseteq X$.

Then:

$A^- + B^- \subseteq \paren {A + B}^-$

where $A^-$, $B^-$ and $\paren {A + B}^-$ denote the closures of $A$, $B$ and $A + B$.

Proof 1

Let $a \in A^-$ and $b \in B^-$.

We aim to show that $a + b \in \paren {A + B}^-$.

It suffices to show that for each open neighborhood $W$ of $a + b$ we have:

$W \cap \paren {A + B} \ne \O$

Fix an open neighborhood $W$ of $a + b$.

From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods, there exists an open neighborhood $W_1$ of $a$ and an open neighborhood $W_2$ of $b$ such that:

$W_1 + W_2 \subseteq W$

Since $W_1$ is an open neighborhood of $a$ and $a \in A^-$ we have:

$W_1 \cap A \ne \O$

Since $W_2$ is an open neighborhood of $b$ and $b \in B^-$ we have:

$W_2 \cap B \ne \O$

Let:

$x \in W_1 \cap A$

and:

$y \in W_2 \cap B$

Then we have $x \in W_1$ and $y \in W_2$ so that:

$x + y \in W_1 + W_2$

Since $W_1 + W_2 \subseteq W$, we then obtain $x + y \in W$.

Since we also have $x \in W_1$ and $y \in W_2$, we have $x + y \in A + B$.

So:

$x + y \in W \cap \paren {A + B}$.

We can therefore conclude that:

$W \cap \paren {A + B} \ne \O$

Since $W$ was arbitrary, we have that $a + b \in \paren {A + B}^-$ as required.

$\blacksquare$

Proof 2

Let $x \in A^-$ and $y \in B^-$.

From Point in Set Closure iff Limit of Net, there exists directed sets $\struct {\Lambda_1, \preceq_1}$ and $\struct {\Lambda_2, \preceq_2}$ and nets $\sequence {x_\lambda}_{\lambda \mathop \in \Lambda_1}$ valued in $A$ and $\sequence {y_\lambda}_{\lambda \mathop \in \Lambda_2}$ valued in $B$ such that:

$\sequence {x_\lambda}_{\lambda \mathop \in \Lambda_1}$ converges to $x$

$\sequence {y_\lambda}_{\lambda \mathop \in \Lambda_2}$ converges to $y$.

Define a relation $\sqsubseteq$ on $\Lambda_1 \times \Lambda_2$ by:

$\tuple {\lambda_1, \lambda_2} \sqsubseteq \tuple {\mu_1, \mu_2}$

for $\tuple {\lambda_1, \lambda_2}, \tuple {\mu_1, \mu_2} \in \Lambda_1 \times \Lambda_2$ if and only if:

$\lambda_1 \preceq_1 \mu_1$ and $\lambda_2 \preceq_2 \mu_2$

From Product of Directed Sets is Directed Set, $\struct {\Lambda_1 \times \Lambda_2, \sqsubseteq}$ is a directed set.

Define:

$x'_{\tuple {\lambda, \mu} } = x_\lambda$

and:

$y'_{\tuple {\lambda, \mu} } = y_\mu$

for each $\tuple {\lambda, \mu} \in \Lambda_1 \times \Lambda_2$.

We show that $\langle {x'_{\tuple {\lambda, \mu} } \rangle}_{\tuple {\lambda, \mu} \mathop \in \Lambda_1 \times \Lambda_2}$ converges to $x$.

Fix $\mu_0 \in \Lambda_2$.

Let $U$ be an open neighborhood of $x$.

Since $\sequence {x_\lambda}_{\lambda \mathop \in \Lambda_1}$, there exists $\lambda_0 \in \Lambda$ such that for all $\lambda \in \Lambda$ with $\lambda \succeq \lambda_0$, we have $x_\lambda \in U$.

Then, if $\tuple {\lambda, \mu} \in \Lambda_1 \times \Lambda_2$ has $\tuple {\lambda, \mu} \sqsupseteq \tuple {\lambda_0, \mu_0}$, we in particular have $\lambda \succeq \lambda_0$ and so $x'_{\tuple {\lambda, \mu} } \in U$.

Since $U$ was arbitrary, $\sequence {x'_{\tuple {\lambda, \mu} } }_{\tuple {\lambda, \mu} \mathop \in \Lambda_1 \times \Lambda_2}$ converges to $x$.

Similarly, $\sequence {y'_{\tuple {\lambda, \mu} } }_{\tuple {\lambda, \mu} \mathop \in \Lambda_1 \times \Lambda_2}$ converges to $y$.

From Sum of Convergent Nets in Topological Vector Space is Convergent, $\sequence {x'_{\tuple {\lambda, \mu} } + y'_{\tuple {\lambda, \mu} } }_{\tuple {\lambda, \mu} \mathop \in \Lambda_1 \times \Lambda_2}$ converges to $x + y$.

Since $x'_{\tuple {\lambda, \mu} } \in A$ and $y'_{\tuple {\lambda, \mu} } \in B$ for all $\tuple {\lambda, \mu} \in \Lambda_1 \times \Lambda_2$, we have that $x'_{\tuple {\lambda, \mu} } + y'_{\tuple {\lambda, \mu} } \in A + B$ for each $\tuple {\lambda, \mu} \in \Lambda_1 \times \Lambda_2$.

So $x + y$ is the limit of a net valued in $A + B$.

So by Point in Set Closure iff Limit of Net, we have $x + y \in \paren {A + B}^-$.

So, we have $A^- + B^- \subseteq \paren {A + B}^-$.

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.13$: Theorem