# Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval/Lemma 2/Lemma 2.1

## Theorem

Let $I = \closedint a b$.

Let $\map \CC I$ be the set of continuous functions on $I$.

Let $\map \DD I$ be the set of continuous functions on $I$ that are differentiable at a point.

Let $d$ be the metric induced by the supremum norm.

Let:

$\ds A_{n, \, m} = \set {f \in \map \CC I: \text {there exists } x \in I \text { such that } \size {\frac {\map f t - \map f x} {t - x} } \le n \text { for all } t \text { with } 0 < \size {t - x} < \frac 1 m}$

Then:

for each $\tuple {n, \, m} \in \N^2$, $A_{n, \, m}$ is closed in $\tuple {\map \CC I, d}$.

## Proof

Fix $\tuple {n, \, m} \in \N^2$.

$\tuple {\map \CC I, d}$ is complete.
$A_{n, \, m}$ is closed if and only if $\tuple {A_{n, \, m}, d}$ is complete.

Let $\sequence {f_i}_{i \in \N}$ be a Cauchy sequence in $\tuple {A_{n, \, m}, d}$.

Since $\tuple {\map \CC I, d}$ is complete, $\sequence {f_i}_{i \in \N}$ converges some $f \in \map \CC I$.

We aim to show that $f \in A_{n, \, m}$.

Since $f_i \in A_{n, \, m}$ for each $i \in \N$, there exists $x_i \in I$ such that:

$\ds \size {\frac {\map {f_i} t - \map {f_i} {x_i} } {t - x_i} } \le n$ for each $t$ with $0 < \size {t - x_i} < \frac 1 m$.

Note that since $I$ is bounded, $\sequence {x_i}_{i \in \N}$ is bounded.

Therefore, by the Bolzano-Weierstrass Theorem:

there exists a convergent subsequence of $\sequence {x_i}_{i \in \N}$, $\sequence {x_{i_k} }_{k \in \N}$.

Let:

$\ds x = \lim_{k \mathop \to \infty} x_{i_k}$

Note that we also have:

$\ds f = \lim_{k \mathop \to \infty} f_{i_k}$

From Subset of Metric Space contains Limits of Sequences iff Closed, since $I$ is closed, $x \in I$.

the sequence $\sequence {\map {f_{i_k} } {x_{i_k} } }_{k \in \N}$ converges to $\map f x$.

We therefore have:

 $\ds \size {\frac {\map f t - \map f x} {t - x} }$ $=$ $\ds \lim_{k \mathop \to \infty} \size {\frac {\map {f_{i_k} } t - \map {f_{i_k} } {x_{i_k} } } {t - x_{i_k} } }$ $\ds$ $\le$ $\ds n$

for all $t$ with $0 < \size {t - x} < \frac 1 m$.

That is, $f \in A_{n, \, m}$.

$\blacksquare$