# Baire Category Theorem and nowhere differentiable function

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 $$0$$?

• You cannot show that this set is closed. Take a look at your source again and see which set is supposed to be closed. – Kavi Rama Murthy Apr 27 at 23:20
• @KaviRamaMurthy What is the set? – gtoques Apr 27 at 23:21
• @KaviRamaMurthy Is it "for any h" instead of just small enough h? – gtoques Apr 27 at 23:22
• 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. – Kavi Rama Murthy Apr 27 at 23:23
• @KaviRamaMurthy So is the idea to do it for smaller and smaller h and then intersect all those sets to get arbitrarily small h? – gtoques Apr 27 at 23:27

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 $$|f(t+h)-f(t)| \leq n|h|\}$$?