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As seen here, standard application of the Baire Category Theorem is to show that the set of functions differentiable at a point in $[0,1]$ is a nowhere dense set in the space of continuous functions on $[0,1]$. For that we construct sets $$ C_{n} = \{f \in C[0,1] : \exists a, \delta \in [0,1] : \frac{|f(a+h) - f(a)|}{|h|} \leq n \text{ for all } |h| < \delta\} $$ The standard proof shows that each $C_{n}$ is closed and has empty interior. I get the empty interior bit, but I'm confused about why it's closed. Yes, we take a sequence $f_{k}$ in each $C_{n}$ and find a point $x_{k}$ for each $f_{k}$ where the above property holds and find a convergent subsequence of $\{x_{k}\}$. My question: how is taking a limit of the $x_{k}$ sufficient? Don't we need to take care of $\delta$ as well? When we take $k \to \infty$, why can we fix $h$? Each $f_{k}$ might have a different value of $h$ that works, so even the $h$ has to be part of our sequence. But we introduce a sequence $h_{k}$, how do we take care of the $\delta$ for each $h_{k}$? Who's to say that $h_{k}$ doesn't converge to $0$?

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  • $\begingroup$ You cannot show that this set is closed. Take a look at your source again and see which set is supposed to be closed. $\endgroup$ – Kavi Rama Murthy Apr 27 at 23:20
  • $\begingroup$ @KaviRamaMurthy What is the set? $\endgroup$ – gtoques Apr 27 at 23:21
  • $\begingroup$ @KaviRamaMurthy Is it "for any h" instead of just small enough h? $\endgroup$ – gtoques Apr 27 at 23:22
  • $\begingroup$ Go to the link you have given and write down the complement of $A_n$ corre tly. Then use the fact that intersection of closed sets is closed. This allows you to prove closedness for a fixed $h$ which is easy. $\endgroup$ – Kavi Rama Murthy Apr 27 at 23:23
  • $\begingroup$ @KaviRamaMurthy So is the idea to do it for smaller and smaller h and then intersect all those sets to get arbitrarily small h? $\endgroup$ – gtoques Apr 27 at 23:27
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Let $A_n$ be the set in your link. Then $A_n^{c}=\cap_h \{f:\exists t: |f(t+h)-f(t)| \leq n|h|\}$. Fix $h$. If $(f_j)$ is a sequence in $\{f:\exists t: |f(t+h)-f(t)| \leq n|h|\}$ converging uniformly to $f$ then $|f_j(t_j+h)-f(t_j)| \leq n|h|$ and there is a subsequence of $(t_j)$ which converges to some $t$. Can you show that $|f(t+h)-f(t)| \leq n|h|\}$?

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