Let X be a topological space and let Y be a metric space

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seed21
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Hello.

Please, help me with this exercise:

Let X be a topological space and let Y be a metric space. Let [tex]f_n: X \rightarrow Y[/tex] be a sequence of continuos functions. Let [tex]x_n[/tex] be a sequence of points of X converging to x. Show that if the sequence [tex](f_n)[/tex] converges uniformly to [tex]f[/tex] then [tex](f_n(x_n))[/tex] converges to f(x).

Thanks in advance.
 
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Okay, here it is.
I don't know if this is correct...

Let [tex]A \in \tau[/tex] such that [tex]f(x)[/tex] [tex]\in A[/tex], then, [tex]f^-^1(A)[/tex] [tex]\in \tau[/tex] and [tex]x[/tex] [tex]\in[/tex] [tex]f^-^1(A)[/tex]

Since [tex]x_n[/tex] [tex]\rightarrow[/tex] [tex]x[/tex] implies that [tex]\exists k \in \mathbb N[/tex] such that [tex]x_n[/tex] [tex]\in[/tex] [tex]f^-^1(A)[/tex] [tex]\forall[/tex] [tex]n \succ k[/tex]



or


given [tex]\epsilon \succ 0[/tex], [tex]\exists k \in \mathbb N[/tex] such that [tex]d(x_n, x)[/tex] [tex]\prec \epsilon[/tex]


Besides, [tex](f_n)[/tex] converges uniformly to [tex]f[/tex] i.e. [tex](f_n)\stackrel{u}{\rightarrow} f[/tex], implies that given [tex]\epsilon \succ 0[/tex], [tex]\exists k \in \mathbb N[/tex] such that [tex]d(f_n(x), f(x))[/tex] [tex]\prec \epsilon[/tex] [tex]\forall[/tex] [tex]n \succ k[/tex]



I think, I have to show that exists [tex]m \in \mathbb N[/tex] such that [tex]f_n(x_n)[/tex] [tex]\in A[/tex] [tex]\forall[/tex] [tex]n \succ m[\tex]<br /> <br /> I don't know how to do that.<br /> I need help.<br /> This is all I have. Thanks in advance.[/tex]