In the linked question: Holder Continuous Functions on $[0,1]$ are complete + Banach space OP is trying to show an $\alpha$-Holder space ($\Lambda_{\alpha}$) is a Banach space. The first answer reads proceeds in three steps: producing a limit $f$, showing that $f_n \rightarrow f$ in norm, and finally showing $f$ is in the space.

To produce $f$ the poster assumes there's a Cauchy sequence in $\Lambda_{\alpha}$ and then states there exists a pointwise limit $f$ of this cauchy sequence.

My question: If we show $f \in \Lambda_{\alpha}$ aren't we done? -- we've assumed that a cauchy sequence in $\Lambda_{\alpha}$ and proved the limit lives in $\Lambda_{\alpha}$. If so, why does the poster bother checking $f_n \rightarrow f$ in norm separately in step 1? They should be able to just skip to step 2, right?

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    $\begingroup$ Because before proving that, we don't know if $f$ is the limit in the norm sense, and not just a limit in the pointwise sense. The completeness is a statement about the convergence in a particular topology, which is here the norm topology. The fact that the pointwise limit is in the same normed space of all the $f_n$ doesn't imply that the convergence is in the norm sense. $\endgroup$ – Phil-W Nov 18 '18 at 22:30
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    $\begingroup$ Indeed, consider the sequence of functions $f_n$ defined by $x \mapsto x^n$ in the space of bounded continuous functions on the halfopen interval $[0,1)$ equipped with the supremum norm. The pointwise limit is the zero function which certainly lives in this space, but the norm distance between $f_n$ and $0$ is $1$ for any $n$. $\endgroup$ – Thomas Bakx Nov 18 '18 at 22:40
  • $\begingroup$ @Phil-W feel free to add your comment as an answer, I can accept it $\endgroup$ – yoshi Nov 18 '18 at 23:31

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