# Why is this definition of convergence incorrect?

From my understanding, a sequence $$\{a_{n}\}$$ is said to converge to $$a$$ if for all $$\epsilon > 0$$, there exists an index $$N$$ such that for all $$n \geq N$$, we have

$$|a_{n} - a| < \epsilon.$$

But, why is this different from the following definition:

There exists an index $$N$$ for every $$\epsilon > 0$$, such that for all $$n \geq N$$, we have

$$|a_{n} - a| < \epsilon$$

Pretty much, I switched the $$\forall \epsilon > 0$$ quantifier with the $$\exists N$$ quantifier, and the definition becomes invalid, but why?

• think of $\lbrace 1/n \rbrace$, you know that it converges to $0$. On the other hand exchanging the quntifiers for convergence you would have to find an $N$ for which $\mid 1/n \mid < \epsilon$ $\forall \epsilon>0$, $\forall n>N$ which is simply not possible. – Baol Oct 1 '18 at 10:39

I am not sure why the other current answers are interpreting your second definition as invalid, because as a native English speaker I interpret it to mean exactly the same (logically speaking) as the first definition. In particular, you did not switch the quantifiers, even though the surface text appears to have them switched. You wrote:

There exists an index $$N$$ for every $$\epsilon > 0$$, such that for all $$n \geq N$$, we have ...

This means that there is some $$N$$ for every $$\epsilon > 0$$, not necessarily the same $$N$$ for all $$\epsilon > 0$$. So it conveys the same logical structure as the other definition, whereby for any given $$\epsilon > 0$$ there is an index $$N$$ such that ...

It would become invalid if you had the following English phrasing (which switches the quantifiers):

There exists an index $$N$$ such that, for every $$\epsilon > 0$$ and for all $$n \geq N$$, we have ...

But you did not use that phrasing, so that does not imply anything about yours.

Just for fun, here is a quote (apocryphally attributed to Albert Einstein) using exactly that phrase structure:

Stay away from negative people. They have a problem for every solution.

It is clearly understood by everyone to mean:

Stay away from negative people, because for every solution they will find some problem with it.

• By the way, I am not in favour of using this kind of phrasing as in "There exists an index N for every ϵ>0", because it is not in line with the logical structure. However, it is not wrong, and one should not call it wrong just because it does not match up nicely with the proper logical structure of "$\forall ε>0\ \exists N \in \mathbb{N}\ \forall n \in \mathbb{N}_{<n}\ ( |a_n-a|<ε )$". – user21820 Oct 1 '18 at 14:37
• @CarstenS: That's a good question! I think that a normal English speaker would understand the phrase "there is an X for every/each Y" correctly, as I cannot think of a situation where it is likely to be misunderstood, whether using "every" or "each". Of course, in other English phrases "every" and "each" can convey really different things. English is really a mess! =) – user21820 Oct 1 '18 at 14:46
• +1 for the overall explanation but more importantly for the advice from Einstein. – Paramanand Singh Oct 1 '18 at 15:08
• Source for that Einstein "quote"? I recently saw that attributed to Feynman, and doubt it is by either of them. "On the internet, everyone puts random names on random quotes" (Abraham Lincoln). – Torsten Schoeneberg Oct 1 '18 at 22:09
• @TorstenSchoeneberg: I had not fact-checked the source of that quote, though I didn't quite expect it to be made-up. In any case, someone has edited my post to say "apocryphally attributed to Einstein", so that's good. =) – user21820 Oct 2 '18 at 8:19

I think, based on your wording and the answers / comments, that there is a misunderstanding in the way things are written / thought.

There is a difference between

$$\exists N$$ such that $$\forall \epsilon>0$$, such that $$\forall n>N$$, $$|a_{n} - a| < \epsilon$$

and

$$\exists N_\epsilon$$ for all $$\epsilon>0$$, such that $$\forall n>N_\epsilon$$, $$|a_{n} - a| < \epsilon$$

You wrote the first (it is at least implied), hence the answers you got. I believe you were meaning the second. The second is, roughly, equivalent to the initial definition you are mentionning. But it is a bit of an awkward formulation, open to misunderstanding (as you can see), even with proper syntax.

• To make it more explicit: If $N$ depends on $\epsilon$, then $\epsilon$ should come first in the definition. By having $N$ come first, you are claiming that it is independent of $\epsilon$. – Teepeemm Oct 1 '18 at 14:11
• @Teepeemm: No; English does not work that way. When we say "there is a carpet for every room", the order obviously does not imply that there is a single carpet that is for all the rooms! Please see my answer. – user21820 Oct 1 '18 at 14:26
• @user21820 English does not work that way; mathematical definitions do (at least if we want to have the phrasing match the logical structure). That's why I like your answer the best. Even though OP has rearranged the English, the logical structure remains the same. But if we want a clear definition, then the English should match the logical structure. – Teepeemm Oct 1 '18 at 20:57
• Using logical quantifiers to act as shorthand notation for English, like in this answer, is very confusing and not required unless you're short on space. It makes it look like you're using the syntax and semantics of the langauge for which the quantifiers exist which causes misinterpretations in the order. See also math.stackexchange.com/a/2933538/98449 – JiK Oct 1 '18 at 21:55
• @Teepeemm: Yes! Hence my comment below my answer, because I sure didn't want to give people the impression that I don't have an issue with using unusual natural language features to convey something that could have been much more straightforwardly conveyed. However, English in a mathematical writing often varies and is not as tightly bound to the logical structure, and so one still has to learn how to interpret it as the language permits. For example, "There is a 4-colouring of every planar graph." and "There is a choice-function on every set of non-empty sets.". – user21820 Oct 2 '18 at 8:31

Look at your second definition and take a $$N$$ such that for every $$\varepsilon > 0$$, $$\forall n \geq N$$, $$|a_n-a| \leq \varepsilon$$.

This means that for all $$n \geq N$$, $$|a_n - a|$$ is as small as you want. So it has to be zero. So your definition is equivalent to $$\exists N \in \mathbb{N}, \forall n\geq N, a_n = a$$

i.e. $$(a_n)$$ is stationary. Of course this is not equivalent to the convergence.

• Or $\forall x \text{ s.t.} |x-1| \geq \delta, |f(x) - f(1)| \geq \epsilon$ – Acccumulation Oct 1 '18 at 16:47