The given set is closed in $\mathbb{R}$? Let $K$ is a non empty closed subset of $\mathbb{R}$ , then show that the set {${x+y : x\in K , y\in [1,2]}$} is closed in $\mathbb{R}$.
I know a set will be closed if it contains all it's limit points but not know how to apply this here...
 A: Let $(z_n)_{n \in \mathbb{N}}$ be a convergent sequence in $\{x + y: x \in K, y \in [1, 2]\}$ with limit $z$. By definition for each $z_n$ there exist $x_n \in K$ and $y_n \in [1, 2]$ such that $z_n = x_n + y_n$. Since $[1, 2]$ is compact by Heine-Borel, there must exists a convergent subsequence $(y_{n_k})_{k \in \mathbb{N}}$ such that $y_{n_k} \to y \in [1, 2]$. But then $x_{n_k} \to z - y$, i.e. we have a convergent subsequence $(x_{n_k})_{k \in \mathbb{N}}$ in $K$. But since $K$ is closed, we have that $z - y \in K$. 
But now note that $z = (z - y) + y$, where $(z-y) \in K$ and $y \in [1, 2]$. That is, the limit of $(z_n)_{n \in \mathbb{N}}$ is in the form $x + y: x \in K$ and $y \in [1, 2]$, and thus $\{x + y: x \in K, y \in [1, 2]\}$ is closed.
A: Suppose $z_n$ is a sequence in $S=\{x+y: x\in K, y \in [1,2]\}$ such that $z_n\rightarrow z$. You have to show $z\in S$.
Since $ z\in S$ there are $x_n,y_n$ such that $x_n\in K$ and $ y_n\in [1,2]$ and $z_n = x_n+y_n$. Since $[1,2]$ is closed and bounded, a subsequence $y_{n_k}$ of $y_n$ converges.
Can you finish from here?
