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So I was looking at a proof that shows that the metric space $(C[0,1],d_\infty)$ is complete.

I know that the definition of complete is that given a metric space X , every cauchy sequence in X converges to a point in X.

The proof was broken into 3 parts

1.create a candidate function and show that the sequence of functions $\{f_n\}$ is cauchy and that the limit $\lim_{n \rightarrow \infty}f_n(a)$ exists.

2.show f is continuous

3.show that $d_\infty(f_n,f)\rightarrow 0$

I understand why we do 1. It's so we can define a cauchy sequence which converges and 3. because we then show that it converges to a point in our space. But I'm not totally sure why we need to show that f is continuous , could someone please explain ?

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  • $\begingroup$ Every Cauchy sequences converges to a point in $\mathbf X$. You need to show that $f$ ,which is your candidate for the limit, is in $X$, so you need to show it is continuous. $\endgroup$ – астон вілла олоф мэллбэрг Aug 9 '18 at 14:33
  • $\begingroup$ @астонвіллаолофмэллбэрг oh of course , because otherwise it wouldn't be a member of $C[0,1]$, right ? $\endgroup$ – excalibirr Aug 9 '18 at 14:34
  • $\begingroup$ Exactly. You can show that such $f$ exists, but if it were not continuous then it would not be in $X$, and you would not have shown completeness. $\endgroup$ – астон вілла олоф мэллбэрг Aug 9 '18 at 14:35
  • $\begingroup$ @астонвіллаолофмэллбэрг Thank you, it seems obvious now that you say it XD $\endgroup$ – excalibirr Aug 9 '18 at 14:38
  • $\begingroup$ You are welcome. For completeness, I will add an answer below, $\endgroup$ – астон вілла олоф мэллбэрг Aug 9 '18 at 14:39
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Note that for completeness you require that your candidate limit $f$ be in $X$. For this, you must show that it is continuous, for otherwise you cannot assert that the limit is in $X = \mathbf C[0,1]$. This is why the second step is important.

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