Convergence towards a point outside a metric space

Question

In my textbook it says a sequence $\{x_i\}$ converges to $x\in M$ if for every $\epsilon>0$ there exists a positive integer $N_\epsilon$ such that $D(x_n,x)<\epsilon$ for all $n>N_\epsilon$.

Suppose we have a metric space $M=(0,1]$. Is it then wrong to say that the sequence $\{x_n\}=\{1,\frac12, \frac13,\frac14,...\}$ is convergent, since $0\not\in M$?

I mean, as $n\rightarrow \infty$, $x_n$ gets arbitrarily close to $0$ and I'd say that means it converges to $0$. On the other hand, the definition seems to require that the limit is contained within $M$ in order to say it converges.

Answer 1

The sequence converges to $0$ in the reals, $0 \notin M$, so $D(0, \frac{1}{n})$ is not even defined! So the sequence does not converge in $M$.

A metric is a function $D: M \times M \rightarrow \mathbb{R}$, so distance is only defined for pairs where both are in $M$.

But it can be the case that in a larger metric space $(X,d)$ we have a subset $M$ that gets the restricted metric $d_M=d|_{(M \times M)}$, as in your case.

Then $x_n \rightarrow x$ in $(X,d)$ implies $(x_n) \rightarrow x$ in $(M ,d_M)$ iff $\forall n : x_n \in M \land x \in M$. If then however $x \notin M$, we have found that $M$ is not closed in $X$: we can "walk out of" $M$ with a sequence; $x$ is then a point not in $M$ that has points in $M$ lying as close as we like to it. Hence $x \in \overline{M}\setminus M$.

Answer 2

There is difference between "a sequence approaches the limit l" and "a sequence is convergent".

When we say a sequence approaches the limit l, it means two things:(a) the sequence tends to the limit l when n tends to infinity and (b) the limit l is defined. If any one of these two fails, we say the sequence has no limit e.g. sequence {1/n} approaches/tends the limit $0$, it has a limit $0$ but the sequence $(-1)^n$ does not approach any limit, its limit does not exist, it has no limit, although its only two subsequences have their own limits 1 and -1. Sequence {n} has no limit in $R$ because its limit is not defined in $R$

Note that: Limit and limit point are two different things.

When we say a sequence is convergent, it means two things: (a) it has a limit (it approaches a limit) and (b) that limit is in the metric space. If any one of these two fails, then the sequence is not convergent. e.g. sequence $(-1)^n$ is not convergent in $R$ because it has no limit. Sequence {1/n} is not convergent in the space $(0,1)$ because it has the limit $0$ but $0$ is not in the space. {1/n} is convergent in $R$ because it has the limit $0$ and $0$ is in the space $R$