Real Analysis – Part 24 – Pointwise Convergence
Real Analysis – Part 24 – Pointwise Convergence

I am studying Real analysis using the book of Axler (Measure Integration and Real analysis) in which the author gives an example about a function that converges pointwise but not uniformly as follows:

However, it seems to me that it is possible to prove that $f_k$ converges uniformly to $f$ as follows:

+) If $x \neq 0$
Given $\epsilon > 0$, we have $|f_k(x) – f(x)| = |f_k(x) – 1|$
For every $x \neq 0$, there exists a natural number $m$ such that $m > \frac{1}{x}$ (i.e. $x \gt \frac{1}{m}$).
With $k \geq m$, $f_k(x) = 1$ and so $|f_k(x) – f(x)| = |f_k(x) – 1| = |1-1| = 0 \lt \epsilon$ for all $k \geq m$

+) If $x = 0$, given $\epsilon > 0$, we have |$f_k(0) – f(0)$| = $0 \lt \epsilon$ for all $k$
So that for all $k \geq m$ ($m$ is taken from above), we have |$f_k(0) – f(0)$| = $0 \lt \epsilon$

To conclude, for every $\epsilon \gt 0$, there exists a natural number $m$ such that $|f_k(x) – f(x)| \lt \epsilon$ for all $k \geq m$ and for all $x \in X$.

As for me, the proof makes sense which contradict the claim of the author. Could you please help me explain this issue ?

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