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In a book I am reading, they mention the following as an example of a Cauchy sequence which is not convergent:
Consider the set of all bounded continuous real functions defined on the closed unit interval, and let the metric of the set be
d(f,g)=[tex]\int_0^1 \! |f(x)-g(x)| \, dx[/tex].
Let [tex](f_n)[/tex] be a sequence in this space defined as:
[tex]f_n(x)=1[/tex] if [tex]0 \leq x \leq 1/2[/tex]
[tex]f_n(x)=(-2)^n(x-1/2)+1[/tex] if [tex]1/2 \leq x \leq 1/2+(1/2)^n[/tex]
[tex]f_n(x)=0[/tex] if [tex]1/2+(1/2)^n\leq x \leq 1[/tex]
I can see that this is a Cauchy sequence, but I can't see how this sequence does not converge. I would say that it converges to:
[tex]f_n(x)=1[/tex] if [tex]0 \leq x \leq 1/2[/tex]
[tex]f_n(x)=0[/tex] if [tex]1/2 < x \leq 1[/tex]
I would appreciate if anybody can help me to see why this sequence does not converge.
Consider the set of all bounded continuous real functions defined on the closed unit interval, and let the metric of the set be
d(f,g)=[tex]\int_0^1 \! |f(x)-g(x)| \, dx[/tex].
Let [tex](f_n)[/tex] be a sequence in this space defined as:
[tex]f_n(x)=1[/tex] if [tex]0 \leq x \leq 1/2[/tex]
[tex]f_n(x)=(-2)^n(x-1/2)+1[/tex] if [tex]1/2 \leq x \leq 1/2+(1/2)^n[/tex]
[tex]f_n(x)=0[/tex] if [tex]1/2+(1/2)^n\leq x \leq 1[/tex]
I can see that this is a Cauchy sequence, but I can't see how this sequence does not converge. I would say that it converges to:
[tex]f_n(x)=1[/tex] if [tex]0 \leq x \leq 1/2[/tex]
[tex]f_n(x)=0[/tex] if [tex]1/2 < x \leq 1[/tex]
I would appreciate if anybody can help me to see why this sequence does not converge.