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Let $X$ be a topological space and $Y$ a metric space. Let $f_n \colon X \to Y$ be a sequence of continuous functions. Let $x_n$ be a sequence of points of $X$ converging to a point $x \in X$. Suppose that the sequence $(f_n)$ converges uniformly on $X$ to a function $f \colon X \to Y$.

Then how to prove that the sequence $(f_n(x_n))$ converges to $f(x)$ with respect to the metric on $Y$?

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You can use the inequality $d(f_n(x_n),f(x))\leq d(f_n(x_n),f(x_n))+d(f(x_n),f(x))$.

One term goes to $0$ because $f_n\to f$ uniformly. The other term goes to $0$ because $f$ is continuous (which you can also prove if you haven't yet).

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  • $\begingroup$ But you see, the first term on the right involves $x_n$, which will also potentially vary as $n$ goes to infinity. How can we then justify the limit approaching $0$ in the first term on the right side of the inequality you have given? $\endgroup$ – Saaqib Mahmood Feb 10 '13 at 9:28
  • $\begingroup$ @Saaqib: For the first term, use uniform convergence. Directly apply the definition and see what it tells you about $d(f_n(x_n),f(x_n))$. $\endgroup$ – Jonas Meyer Feb 10 '13 at 11:01
  • $\begingroup$ Yes, it's clear now. Thank you so much! Visually impaired, I'm using a different sort of display scheme and so can't see all the icons; hence my inability to rate/accept answers. $\endgroup$ – Saaqib Mahmood Feb 11 '13 at 9:41

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