# limit in open set in metric space

I have started studying metric spaces by my own. I am getting problem with the concept of closed set. I am reading from Irving Kaplansky. This book has started the concept of closed set in relation with limit of convergent sequence in a set and according to which closed set in metric space are the ones which contains all limits of convergent sequence in it. I can understand this a bit.But I think same thing is also true for open set in a metric space.Here is why I think so:

Let U be an open set in a metric space M.Consider there is a convergent sequence $\{x_n\}_{n=0}^{\infty}$ such that $x=\lim_{n\to\infty} x_n$ and $x$ does not belong toU .So $\ x=\lim_{n\to\infty} x_n$ implies

for each $m>0$ there exists some n in $N$ s.t. $d(x,x_i)<m$ for all $i>=n$

that means if $d(x,x_t)<m$ if $t>n$ ...(1)

$x_t$ lies in U which implies that there exist an open ball with centre at $x_t$ and radius $r$ in U since $m$ can be chosen arbitrarily, we can restate (1) as:

$d(x,x_t)<r$ if $t>n$ i.e. $x$ lies in the open ball of $x_t$ and hence $x$ lies in $U$ which is contradiction.

• The radius depends on the point. Take $U = (0,3)$ and $x_n = \frac{1}{n}$ as a simple example. Commented Nov 10, 2016 at 14:24

The problem is that you do too many things at a time.

The following is still correct: every $x_t$ is an element of $U$ indeed. And since $U$ is an open set, this means that there is an open ball with some radius $r$ around $x_t$ that is completely contained in $U$. So, we have $$\text{if}\quad d(y,x_t)<r \qquad \text{then}\qquad y\in U.$$ Now you feed your found radius into the convergence statement, so you get a number $N$ such that $$\text{if}\quad n>N \qquad \text{then} \qquad d(x,x_n)<r.$$

The problem is now that you do not know anything about $N$. In particular, you do not know if $t>N$. So, you do not know if $d(x,x_t)<r$.