Is a 'normally' convergent sequence still convergent in a metric space which barely excludes its 'normal' limit? For example, suppose
$$ x_n = \frac 1n \\ X = (0, 1)$$
Is $x_n$ convergent in $X$?

My guess would be no, since there exists no $x \in X$ which $x_n$ approaches; $x_n$ will eventually surpass any given $\bar x \in X$.  Assuming I'm right, is there a better reason for this to be so?

The upshot is, if its "normal" limit isn't in the space, then the sequence doesn't converge in the space.
It's often important to make such a distinction (between convergence in a space and convergence in some overlying larger space). For instance, your example sequence shows that $(0,1)$ with the usual metric isn't a complete metric space (if you know the sequence is Cauchy), nor is it compact in any equivalent metric (since the sequence has no convergent subsequence).
