Let $V$ be a normed vector space.
1) Let $(b_n)\subseteq V, b_n \to b\in V.$ Show that $B := \{b,b_1,b_2\dots\}$ is closed.
2) Let $(c_n)\subseteq V$ be Cauchy but not convergent (so $V$ is not complete). Show that $C := \{c_1,c_2,\dots\}$ is closed.
1) I know that if $b_n\to b,$ then $b_n$ is Cauchy. Also, if $(x_n)\subseteq B, x_n \to x$, then if $x\not\in B, ||x_n-x|| > 0\,\forall n.$ But how can I use the fact that $b_n$ is convergent to show that $B$ is closed? I think I can use the convergence of $(b_n)$ to show that $\exists \epsilon_0 > 0$ such that $||x_{n_k}-x|| \geq \epsilon_0\,\forall k$, where $(x_{n_k})$ is a subsequence of $(x_n).$
2) Let $(x_n)\subseteq C, x_n \to x \in V.$ We want to show that $x \in C.$ Since $x_n\to x, (x_n)$ is Cauchy. I think I can use the fact that $(c_n)$ is Cauchy to show that if $x\not\in C,$ then $(x_n)$ is not Cauchy, but I'm not sure how to do this.
$(c_n)$ is Cauchy but not convergent implies the image set $S$ of $(c_n)$ has no limit point. So given any $v\in V-S$ we have a nbd $U_v$ of $v$ such that $U_v$ contains only finitely many elements of $(c_n)$. Since normed are Hausdorff, each point of normed space is a closed set. So deleting those finitely many points of $U_v$ which are elements of $(c_n)$ we may assume $U_v\cap S=\emptyset$. Since $v\in V-S$ is arbitrary, we have $V-S=\bigcup_{v\in V-S}U_v$, which is an open set. So we are done.