Let $X,d_X$ and $Y,d_Y$ be metric spaces and $(f_n)_n$ a sequence of (continuous) functions. Does this hold and, more importantly, why?

$$ d_Y\left(\lim_{n\rightarrow +\infty}f_n(x), \lim_{n\rightarrow +\infty}f_n(y)\right) \le \lim_{n\rightarrow +\infty}d_Y\left(f_n(x),f_n(y)\right) $$

I was hoping for an equality but this inequality is fine as well. (If this doesn't hold, does it hold for uniformly continuous functions $f_n$?)

This was part of a proof, but I'm not convinced it's correct yet.

  • 5
    $\begingroup$ $d_Y$ is continuous, hence the result. $\endgroup$
    – copper.hat
    Jun 3, 2015 at 14:57
  • $\begingroup$ @copper.hat, does that require all the limits to exist? $\endgroup$
    – TravisJ
    Jun 3, 2015 at 14:58
  • $\begingroup$ @copper.hat, what does it mean for a metric to be continuous (in symbols). I've only seen continuity for functions in one argument. $\endgroup$ Jun 3, 2015 at 14:59
  • 1
    $\begingroup$ @SydKerckhove It means that $d_Y: Y \times Y \longrightarrow \Bbb{R}$ is continuous. Here $Y \times Y$ is the cartesian product, and it can be given a structure of metric space $(Y \times Y, D)$ where $$D((y,w),(y',w'))=\max \{ d_Y(y,y'), d_Y(w,w') \}$$ $\endgroup$
    – Crostul
    Jun 3, 2015 at 15:04

1 Answer 1


In general, for any metric, you have $d(x,y) \le d(x,z)+d(z,y)$ and so (repeating for $d(x,z)$) we have $|d(x,y)-d(x,z)| \le d(y,z)$.

Hence $|d(x,y)-d(x',y')| \le |d(x,y)-d(x,y')+d(x,y')-d(x',y')|\le d(x,x')+d(y,y')$.

In particular, if $x_n \to x, y_n \to y$, we have $d(x_n,y_n) \to d(x,y)$.


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