Showing a metric space is not complete.

Question

Consider the metric space $$B = \{ f \in C[0,1] : \int_a^b \left| f(x) \right| dx \leq 1\},$$ where $d(f,g) = \int_0^1 \left| f(x) - g(x) \right|dx$.

I'm trying to show that this metric space is not complete. I have proved that the metric space is not totally bounded. I did this by showing that there existed at least one sequence with a subsequence that was not Cauchy.

I'm aware that a metric space is compact $\Leftrightarrow$ Complete + Totally bounded.

Similarly, I know that compactness $\implies$ completeness, but $\neg$(compact) does not imply $\neg$(complete).

Can't seem to get the logic out.

Answer 1

Let $n \in \mathbb{N}$, and define $ f_n(x) = \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}l@{}} 0 & \text{if } 0 \leq x \leq 1/2\\ 2nx - n & \text{if } 1/2 < x \leq (n + 1)/(2n)\\ 1 & \text{if } (n + 1)/(2n) < x \leq 1\\ \end{array}\right.$

Then $\{f_n\}$ is a Cauchy sequence whose limit is $ f(x) = \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}l@{}} 0 & \text{if } 0 \leq x \leq 1/2\\ 1 & \text{if } 1/2 < x \leq 1\\ \end{array}\right.$

because $d(f,f_n) = \int_0^1 |f(x) - f_n(x)| dx = 1/(4n)$.