The sum of two closed subsets of $\mathbb R$ is not a closed set. What is wrong with the following argument? Suppose $X$ and $Y$ are two closed subsets of $\mathbb R$. Let us define $Z=X+Y=\{x+y~|~x\in X,y\in Y\}$
Since, $X$ is closed, $\forall~ x \in X, \exists$ a sequence in  $ X:\{x_n\}$ that converges to $x$.
Similarily $\forall~ y \in Y, \exists$ a sequence in  $ Y:\{y_n\}$ that converges to $y$.
Consider a point $ z=x+y \in X+Y$. Then, there exists $N \in \mathbb N$ such that 
$|z-(x_n+y_n)| = |(x-x_n)+(y-y_n)| \le |x-x_n|+|y-y_n| \le \epsilon$ whenever $n \ge N$
$\implies \forall z \in X+Y, \exists$ a sequence $\in X+Y$ which converges to $z \implies X+Y $ is closed.

What could be the error in this argument? Thanks a lot for your help.

 A: The trouble with the argument is that it doesn't use a correct definition of "closed". You seem to be using "closed" to mean the following:

A set $X$ is closed if for every point $x\in X$, there is some sequence in $X$ converging to $x$.

This is not what closed means - in fact every set satisfies this because the sequence $x,\,x,\,x,\,\ldots$ always converges to $x$. Rather, a closed subset of $\mathbb R$ is defined as follows:

A set $X$ is closed if for every sequence $x_1,x_2,\ldots,\in X$ that converges to some $x$ in $\mathbb R$, we have $x\in X$.

That is: if a sequence in a closed set converges, it converges in the closed set.
Therefore, to show that $X+Y$ is closed, you would have to start your argument as:

Let $z_1,z_2,\ldots$ be a sequence in $X+Y$ that converges to some $z\in \mathbb R$. We wish to show $z\in X+Y$.

You would probably proceed by saying:

Since each $z_i$ is in $X+Y$, we can come up with two sequences $x_1,x_2,\ldots\in X$ and $y_1,y_2,\ldots\in Y$ such that $x_i+y_i=z_i$.

You would, however, then be stuck: if you want to use that $X$ or $Y$ are closed, you would need to show that the sequences $x_i$ and $y_i$ have limits in $\mathbb R$ which does not follow from what the fact that the sum of these sequences has a limit - in fact, these sequences don't even need to be bounded!
The statement turns out to be false - a good way to get at a counterexample is to set $X=\mathbb Z$ and then see if you can use the fact that open intervals are unions of closed ones to somehow make $Y$ a "pile" of closed intervals whose translates stack into open ones.
