Prove that function tends to 0 everywhere in this interval

[lordianed]

Homework Statement

From Spivak: Suppose that $A_{n}$ is, for each natural number $n$, some finite set of numbers in $[0,1]$, and that $A_n$ and $A_m$ have no members in common if $m\neq n$. Define f as follows:

$f(x) = \frac{1}{n}$, if $x \in A_n$
$f(x) = 0$, if $x \notin A_n$ for all $n$​

Prove that $\lim_{x\to a} f(x) = 0$ for all $a$ in $[0,1]$.

Homework Equations

Epsilon-Delta definition of a limit

The Attempt at a Solution

I'm trying to follow micromass' suggestion, so I would be delighted if someone could assist me. My main concern is that I did not really make use of the assumption that $A_n$ and $A_m$ have no members in common if $m\neq n$.

So ask somebody to rip apart your proof completely.

Proof: We first prove the limit tends towards $0$ for all $a$ in the open interval $(0,1)$. So let $\epsilon >0$ be arbitrary, and let $a$ be an arbirtrary element of $(0,1)$. We can find some natural number $n_{\epsilon}$ such that for all $m\in \mathbb{N}$ for which $m\geq n_{\epsilon}$, $\frac{1}{m} < \epsilon$, so there is only a finite number of natural numbers $1,...,n_{\epsilon}-1$ whose reciprocals are greater than $\epsilon$, and therefore finitely many $y\in [0,1]$ for which $|f(y)| > \epsilon$. Hence, we will be able to find a neighbourhood of $a$ which does not contain any of the $y$'s, and hence there exists a $\delta$ such that for all $x$, $0<|x-a|<\delta$ implies $|f(x)|<\epsilon$. For the endpoint $0$and $1$, the right and left hand limits both tend to 0 respectively, and since $f(x) = 0$ for all $x \notin [0,1]$, we can conclude that $\lim_{x\to 0}f(x) = \lim_{x\to 1}f(x) = 0$.

 

[Samy_A]

My main concern is that I did not really make use of the assumption that $A_n$ and $A_m$ have no members in common if $m\neq n$.

That assumption is (implicitly) used in the definition of the function $f$. If $A_n$ and $A_m$ have members in common, $f$ is not well-defined.

 

[lordianed]

If AnA_n and AmA_m have members in common, ff is not well-defined.

Oh right, thanks Samy, I did not catch that

 

[geoffrey159]

Hi lordianed ! This exercise is very much like the previous one you put on PF a few days ago.

Assume that your function converges to a limit $\ell$ as $x\rightarrow a$. Assume that $a \in ]0,1[$ and adapt the definition for $a\in\{0,1\}$.

So you have,

$\forall \epsilon > 0,\ \exists\delta >0\ \forall x\in [0,1] \ ( x\in[ a - \delta , a + \delta ] \Rightarrow |f(x) - l | \le \epsilon)$

You need to show that in $[ a - \delta , a + \delta ]$, there is an element $x_0$ that belongs to none of the $A_n$. In which case $|\ell| = |f(x_0) - \ell | \le \epsilon$. That proves that if the limit exists, it has to be zero.

 

[lordianed]

Hi lordianed ! This exercise is very much like the previous one you put on PF a few days ago.

Assume that your function converges to a limit $\ell$ as $x\rightarrow a$. Assume that $a \in ]0,1[$ and adapt the definition for $a\in\{0,1\}$.

So you have,

$\forall \epsilon > 0,\ \exists\delta >0\ \forall x\in [0,1] \ ( x\in[ a - \delta , a + \delta ] \Rightarrow |f(x) - l | \le \epsilon)$

You need to show that in $[ a - \delta , a + \delta ]$, there is an element $x_0$ that belongs to none of the $A_n$. In which case $|\ell| = |f(x_0) - \ell | \le \epsilon$. That proves that if the limit exists, it has to be zero.

Thanks Geoffrey, I'll try that method too. However I asked to have my proof ripped apart so to say, I already have tried proving it and would love to see people point out any flaws in my argument in the original post.