Suppose we have a pseudo-metric space $(X, \rho).$ Then we can think of all the sequences of points $X^{\Bbb{N}}$ and pseudo-metric $\rho$ gives the notion of convergence to those sequences. On the other hand we can construct a honest metric space $(\widetilde X, \widetilde{\rho})$ called metric identification using $(X, \rho).$ Now we have a natural projection map $X^{\Bbb{N}}\to \widetilde X^{\Bbb{N}}$ given by $(x_n)\mapsto\bigl([x]_n\bigr),$
My question is, how limit points of a sequence behave under this Metric identification?
- Does every limit point of a sequence in $X$ produces a limit point of a sequence in $\widetilde X$?
- In general, which space contains more limit points ? and why?
These questions looks fairly simple and, I guess the pseudo-metric space has more limit points for a given sequence (thinking of $\rho(x,y)=0\,\,\,\forall x,y$).
Can any of you give me a more logical explanation for theses questions? Thank you in advance.
We can show that, if $x$ is a limit point of $(x_n)$, then $[x]$ is a limit point of $([x_n])$. This directly comes from the fact that $\rho(x_n, x) = \overline{\rho}([x_n], [x])$: because the left side is less than some fixed $\varepsilon > 0$ infinitely many times, so too is the right.
The quotient map $x \mapsto [x]$ is surjective, which shows (in the sense of cardinality) that the are more limit points of $(x_n)$ than $([x_n])$. Don't forget, when $x$ is a limit point of $(x_n)$, and $\rho(x, y) = 0$, then $y$ is also a limit point of $(x_n)$, but both become identified with $[x] = [y]$.