# Alternative proof limit definition is equivalent to sequence limit definition (Rudin 4.2)

I was reading the following equivalence (which is my summarized version of writing theorem 4.2 in Rudin in 1 line, look at Appendix in my Q to see exactly what this means):

$$\lim_{x\to p} f(x) = q \iff \lim_{p_n \to p, p_n \neq p } f(p_n) = q$$

what I am interested is in showing the RHS implies the LHS rigorously. I saw Rudin's proof via contrapositive/contradiction. I think I understand it but was a bit surprised that it was via that method rather than a direct method because the statement seems quite natural to me, though there is one step that is tricky to make formal (perhaps the reason a direct proof doesn't exist?).

Basically what we know is the RHS is true, so that every sequence $(p_n)$ in $E$ that converges to $p$, $p_n \neq p$ has $f(p_n)$ converge to $q$. So whenever we are $\delta$ close to $p$ via $n \geq N_{\delta}$, we have that we have $f(p_n)$ is epsilon close to $q$. The issue is that the RHS doesn't actually talk about $\delta's$ but about $N$'s (since its about sequences). But the idea is simple. Choose a $\delta >0$ regardless. Then because every element in the neighbourhood defined by $\delta$ is an element of $E$ it means we can define any sequence we want from those element (or they can be the tail end of some sequence by appending whatever elements we want that are not within $\delta >0$ but that are still in $E$). Thus, since the RHS is in terms of every sequence in $E$ we can guarantee we can cover any neighbourhood for any $\delta$ (because the RHS is in terms of every sequence in $E$, which is the key part for things to work of course). Thus, since we can cover every neighbourhood of size $\delta$ because enough sequences will cover it means that for those terms we have $f(p_n)$ is within $\epsilon$ from $q$, which is what we needed.

I am pretty sure the proof is correct but had a hard time expressing it in a clean way that makes it 100% clear that its correct. Does anyone know how to do this? (essentially write the direct proof in a convincing unambiguous way)?

I noticed that the key part to show is that because we have a property for every sequence that we can actually cover the whole neighbourhood for any $\delta$. Given a $\delta >0$ consider some terms inside it. Then that corresponds to some sequence. Now consider the remaining points, now define another sequence. Keep doing this forever. Then because the RHS guarantees every sequence is covered (regardless if its in the neighbourhood captured by $\delta >0$ or not) it means that any countable sequence we choose form the neighbourhood has some sequence with the properties of the RHS. Since this is true it must mean that we covered the whole neighbourhood. This holds true for any $\delta$, so it doesn't matter which $\delta$ we choose.

Appendix 1:

Let $X$,$Y$ be metric spaces $E \subset X$ and $p \in LP(E)$ (i.e. $p$ is a limit point of $E$). Then:

$$\lim_{x \to p} f(x) = q$$

if and only if:

$$\lim_{n \to \infty} f(p_n) = q$$

for every sequence $\{ p_n \}$ in $E$ such that:

$$p_n \neq p, \lim_{n \to \infty } = p$$

Appendix 2:

To make the question self contained I will write Rudin's definition of limit i.e what:

$$\lim_{x \to p} f(x) = q$$

means:

If there is a point $q \in Y$ with the following property:

$$\forall \epsilon >0, \exists \delta >0: x \in E \text{ if } 0 < d_{X}(x,p) < \delta \implies d_Y(f(x),y) < \epsilon$$ 

You seem to have something backwards. To prove that $\lim_{x\to p} f(x) = q$, you have to show that for every $\epsilon > 0$, there is a $\delta > 0$ such that if $0 < |x - p| < \delta$, then $|f(x) - q| < \epsilon$.

The key thing to note here is that you are given $\epsilon$, and from it, you must find $\delta$. Thus you cannot just "pick a $\delta$". Not every $\delta$ will work. $\delta$ is to some extent a function of $\epsilon$ - though not entirely, since for any $\delta_\epsilon$ that works, any smaller $\delta$ will also work. But still, the value of $\delta$ will depend upon the value of $\epsilon$.

Now where in your argument do you see that dependence? For a given $\epsilon$, how do you determine the $\delta$ that you need? You just pick an arbitrary $\delta$ and start talking about sequences within the $\delta$-neighborhood. Presumably, you mean sequences converging to $p$, since that is all we have any guarantee about, but even with that correction, these sequences do not tell us what we need.

We need to show that for every $x$ inside the $\delta$-neighborhood of $p$, we have $f(x)$ inside the $\epsilon$-neighborhood of $q$. Any given $x$ will belong to infinitely many sequences converging to $p$. But there is no guarantee that on any of those sequences, it will occur late enough that $f(x)$ is within $\epsilon$ of $q$.

To be more explicit, suppose $\{p_n\}$ is a sequence converging to $p$ with $p_K = x$ for some $K$. What we know about this sequence is that $\lim_n f(p_n) = q$. And therefore, there is some $N$ such that for all $n \ge N, |f(p_n) - q| < \epsilon$. But we have no way of knowing if $K \ge N$. Since we don't know that $K \ge N$, we also don't know whether $|f(x) - q| = |f(p_K) - q|$ is less than $\epsilon$.

Indeed, for any $x \in E$, if $f(x) \ne q$, then we know there are $\epsilon$ for which this does not work. if $\epsilon < |f(x) - q|$, then under no sequence $\{p_n\}$ with $p_K = x$ and with $N$ being sufficient to have $|f(p_n) - q| < \epsilon$ for $n \ge N$, will we have that $K \ge N$. The very fact that $\epsilon < |f(x) - q|$ disqualifies it.

So your "proof" does not work.