Let $\mathscr{B}=\{(-a,a)\times \mathbb{R}:a>0\}$. It's easy to see that $\mathscr{B}$ is the filter of neighborhoods of the origin in a topology compatible with the linear structure of $\mathbb{R}^{2}$, namely, $\mathscr{F}(0,0)=\{U \subset \mathbb{R}^{2}: U \supset (-a,a) \times \mathbb{R} \ (\exists a>0) \}$.
My question: How to prove that $E=\{(x,0): x \in \mathbb{R}\}$ is complete?
By definition, a subset $E$ of a topological vector space is said to be complete if every Cauchy filter on $E$ converges to a point $x$ of $E$.
I started considering a Cauchy sequence $(x_n,y_n)$ in $E$. So that, $y_n=0$ por all $n \in \mathbb{N}$. Since $\pi_1(x,y)=x$ is linear and continous, we obtain that $(x_n)$ is a Cauchy sequence in $\mathbb{R}$. Thus, there exists $x \in \mathbb{R}$ such that $x_n \rightarrow x$ as $n \rightarrow \infty$. Now it's easy to see that $(x_n,0) \rightarrow (x,0)$ in the topology above mentioned.
Considering now a Cauchy filter $\mathscr{F}$ in $E$, the uniform continuity of $\pi_1(x,y)=x$ ensures that $\pi_1(\mathscr{F})$ is a Cauchy filter in $\mathbb{R}$. Therefore, there exists $x \in \mathbb{R}$ such that $\pi_1(\mathscr{F}) \rightarrow x$. How could I prove that $\mathscr{F} \rightarrow (x,0)$ in $E$ in the subspace topology?