# How can a Cauchy sequence converge to an irrational number?

I am a physics major and would like to clear a confusion regarding complete metric spaces. I am quoting the definition of a Cauchy sequence from wikipedia below

Formally, given a metric space $(X, d)$, a sequence $x_1, x_2, x_3, \ldots$ is Cauchy, if for every positive real number $\epsilon > 0$ there is a positive integer $N$ such that for all positive integers $m, n > N$, the distance

$$d(x_m, x_n) < \epsilon$$

Now, if we have sequence like $x_1=3, x_2=3.1, x_3=3.14, \ldots$ converging to $\pi$, I do not understand how all distances $d(x_m, x_n)$ will be less than all positive real numbers. Since irrational numbers do not terminate and continue forever, how can the distance ever be less than the smallest real number or infinitesimal (hyperreal) as the distance can never become $0$. Does this definition of completeness apply where $\epsilon$ is infinitesimal (hypperreal) ?

Kindly excuse my ignorance as I am not a mathematics major.

Thanks

-
Don't think of it as one particular distance, $d(x_m,x_n)$, is smaller than every positive real number. But rather, any given positive real number, you can go far enough out to guarantee that $d(x_m,x_n)$ is smaller. – I. Cavey Jan 11 at 3:14
There is no "smallest [positive] real number". – Eric Wofsey Jan 11 at 3:15
It is not that the distance is smaller than every positive real number, it is that for any positive real number, there is a point where the distance is eventually less than it. E.g. can you eventually go far enough that the sequence is accurate to the fifth decimal place? (i.e. $\epsilon=10^{-5}$) Can you go far enough that the sequence is accurate to the hundredth decimal place? (i.e. $\epsilon=10^{-100}$) If I were to give you some number of decimal places, could your sequence eventually be that accurate? The answer of course is yes. – JMoravitz Jan 11 at 3:16
I disagree with the vote to close the question. ${}\qquad{}$ – Michael Hardy Jan 11 at 3:17
@AloizioMacedo : Perhaps it is unclear to you, but not to me. ${}\qquad{}$ – Michael Hardy Jan 11 at 4:02

The subject line currently reads “How can a Cauchy Sequence converge to an irrational number?”.

If we construe that literally, then one easy way a Cauchy sequence (lower-case initial "s") can converge to $\pi$ is that every term of the Cauchy sequence is $\pi$. Thus: $x_1=\pi, x_2=\pi, x_3=\pi,\ldots\,{}$. I suspect you meant “How can a Cauchy sequence of rational numbers converge to an irrational number?”.

Consider your sequence $3,\ 3.1,\ 3.14,\ 3.141,\ \ldots\,$.

The definition DOES NOT say that all distances between members of this sequence are less than all positive numbers. That would happen only with a constant sequence like my first example above. It says:

For every positive real number $\varepsilon>0$ there is a positive integer $N$ such that for all positive integers $m,n>N$ we have $d(x_m,x_n)<\varepsilon$.

Notice that $N$ depends on $\varepsilon$. In fact as $\varepsilon$ gets smaller, typically $N$ must get bigger. Suppose $\varepsilon = 0.01$. Then for your example sequence, $N=3$ is big enough: every pair of numbers in the sequence at or after the third place in the sequence differ from each other by less than $\varepsilon=0.01$. Thus $3.14$ and $3.141$ differ by less than $0.01$. But now suppose $\varepsilon=0.00001$. Then you need a bigger value of $N$. If each term of the sequence has one more digit or $\pi$, then $N=5$ would be big enough for that value of $\varepsilon$.

Notice that the definition of convergence to $\pi$ differs from the definition of "Cauchy sequence". It says for every $\varepsilon>0$ there is a positive integer $N$ such that for every positive integer $n\ge N$ we have $|x_n-\pi|<\varepsilon$. Again, $N$ depends on $\varepsilon$. If $\varepsilon=0.00001$, then $N=5$ would be enough: every term at or beyond the $5$th one differs from $\pi$ by less than $\varepsilon=0.00001$.

There is nothing in either of these definitions that says that the distance between two different members of the sequence or the distance between $\pi$ and a member of the sequence is $0$.

You wrote:

Since irrational numbers do not terminate and continue forever

Let's be clear on a definition.

It is certainly not correct that numbers whose decimal expansions do not terminate are necessarily irrational. For example, $1/7 = 0.\ 142857\ 142857\ 142857\ \ldots$ has a non-terminating decimal expansion and is rational.

Nor is it the case that "rational number" is defined as one whose decimal expansion repeats or terminates. Euclid and other ancient Greeks proved some numbers are irrational without ever thinking about decimal expansions. That $\pi$ is irrational means $\pi$ is not a quotient of two integers, like $22/7$. Proving $\pi$ is irrational is so difficult that it was not done until the 18th century. Some numbers are far easier to prove to be irrational. For example, if $\log_2 3 = m/n$ and $m,n$ are positive integers, then $2^m=3^n$, but that can't happen because an even number cannot be equal to an odd number.

The fact that a number is rational if and only if its decimal expansion repeats or terminates takes a bit of work to prove, but it's elementary enough that high-school students will understand it.

-
In the last sentence, I think you meant "...a number is rational if and only if its decimal expansion repeats or terminates..." – Nathan Reed Jan 11 at 4:34
I don't understand the section after the quoted statement "Since [the decimal expansions of] irrational numbers do not terminate and continue forever." You appear to be trying to rebut that statement, but the statement is true. The asker never claims that numbers whose decimal expansions don't terminate are rational. They also don't claim that the definition of being rational is that the decimal expansion repeats or terminates. Both of those claims would be false but they're never made so I don't understand why you're rebutting them. – David Richerby Jan 11 at 4:55
Agree with @DavidRicherby; this answer would be improved by taking out the rebuttal that infinite decimals are necessarily irrational, which was never part of the OP's inquiry. – Daniel R. Collins Jan 11 at 5:55
Er, I mean the asker doesn't ever claim that non-terminating decimals are irrational. Doh! – David Richerby Jan 11 at 5:57
@NajibIdrissi : Why would you say that? If someone is learning this subject, they should become aware of things like that. "Cauchy sequence" is not defined as meaning a Cauchy sequence of rational numbers. ${}\qquad{}$ – Michael Hardy Jan 13 at 17:08

It's not the case that "all distances $d(x_m, x_n)$ will be smaller than every real number". That is not even true "eventually", in the sense that for some $N$, it's true for all $m, n > N$. That would imply that all such distances are $0$. Furthermore, of course there is no "smallest [positive] real number". The point is that for any positive real number $\epsilon$, no matter how small, the distances between the terms of the sequence eventually get smaller than $\epsilon$ and stay smaller than $\epsilon$.

Think of it as a challenge: I give you some $\epsilon>0$, and you have to find a position in the sequence (some $N$) such that $d(x_m,x_n)<\epsilon$ for all $m,n>N$. If the sequence is Cauchy, you are guaranteed to win the challenge. Cauchy chose the letter $\epsilon$ to stand for error (or erreur): the finite approximations $x_n$ will have an error of less than $\epsilon$ from some point on.

-
In the hyperreal setting mentioned by the OP, the difference can be smaller than every positive real number and still be nonzero. See my answer. – Mikhail Katz Jan 12 at 16:24
The question did not originally mention hyperreals. – BrianO Jan 12 at 16:42
OK, I guess I am coming in late in the game. – Mikhail Katz Jan 12 at 16:42

If $n >m \ge 1$ you have $d(x_n,x_m) < 10^{n-1}$.

Choose $\epsilon>0$ and $N$ such that $10^{N-1} < \epsilon$, then if $n,m \ge N$ we have $d(x_n,x_m) < \epsilon$.

Note that the statement is for any $\epsilon>0$ there is some $N$ such that blah, blah, blah, and not that there is some $N$ such that for all $\epsilon>0$ we have blah, blah, blah.

The latter formulation would imply that the sequence is constant after some $N$.

-

I think your confusion comes from "for every positive real number $\epsilon$". In fact you have to fix $\epsilon$ first when finding $N$.

-

first of all it is not that all the distances are less than every positive number. given a positive number , one can find a stage after which any two terms are the given number close to each other.

The idea of a sequence converging to some point do not necessarily imply that the distance between two terms of the sequence become zero. What it says is that, as $n$ becomes larger, the terms of the convergent sequence starts coming closer to the previous one in terms of distance,

e.g in a sequence $<1/n>$ the distance between 1 and 1/2 is 1/2, distance between 1/2 and 1/4 is 1/4 and so on.. now if you look at distance between any two terms after 1/2, it is always less 1/2, if you look at the terms after 1/100 the max distance between any to terms will be upto 1/100 and so on. So if the given positive number is 1/100, you can choose here $n_0=101$(rather anything bigger than it). whereas if you want distance to be less than 1/1000 say, then this $n_0$ wont work, so you will need $n_{0}=1001$ atleast. It does not depend on terms being rational or irrational.

-

" I do not understand how all distances $d(x_m,x_n)$ will be less than all positive real numbers."

This is the key to your misunderstanding. All distances can't be.

But for any real number no matter who small we can find an infinite number of distances smaller. (So we could say all real numbers are bigger than an infinite number of the distances...)

Let's do an actual example. Consider .9, .99, .999, etc. ($a_n = \sum_{i=1}^n 9/10^i$). This is a cauchy sequence that converges to 1. Okay, 1 is not irrational. But is that really where your misunderstanding lies? Okay, I'll get to that later. But for now.

Cauchy sequence: for any $\epsilon > 0$ ... okay, let's say $\epsilon = 1/1,000,000$ ... we can find a point after which all distances are less than $\epsilon$ ... okay, that'd be simply $n > 7$. $d(.99999\ldots9, .9999\ldots9) < 1/1,000,000$ for any two $.999\ldots99$ with more than 7 digits.

Okay, but let's make $\epsilon$ really small. Let's make it 1/googol. ($10^{-100}$) Well, now if n > 100 we still have $(.999\ldots9, .9999\ldots) < 1/\text{googol}$ if those terms have more that 100 nines. How about a googleplex? Then the terms need a googol nines to be that close. But we can find them with more than a googol nines. No matter how small we can find a point where the following difference of terms is smaller.

That's a cauchy sequence.

Okay, that was a cauchy sequence of rational numbers converging to a rational number. You wanted to know about a cauchy sequence of rational numbers converging to an irrational number and you suggested the decimal expansion of $\pi$. Well, same thing. $\pi$ expanded to a million decimal places will be off but it'll be of by a very small amount. For any real number no matter how small, we can find a point after which all decimal expansions are closer that that to $\pi$. These expansions are still finite but they are long enough to be closer to $\pi$ than the small real number we chose. (And if they aren't, we simply choose the finite decimal expansions that are further out. There'll always be finite expansions further out. That's the point.)

It's true we never get there. And we can't really choose rational values that can get there (that's why the number is irrational). but we've trapped and honed it in with cauchy sequences.

-
thanks for your response. My main confusion has been made clear that you have to choose $\epsilon$ first and then you get $N$. But how does this work with infinitesimal (hyperreals)...How does this definition of completeness apply there ? – phd-applicant Jan 11 at 5:55
hyperreals are outside my area of expertise. My understanding though is that cauchy sequences of reals do not nesc converge to hyperreals. Hyperreals are not archimedian (for any x>0 y you can find integer n s.t. (n-1)x <= y < nx) and so have a different definition of completeness. I really don't know. I'm sorry I don't know more. But the key of the reals is that epsilon of choice is not an infintisimal. – fleablood Jan 11 at 6:22
To say that the fact that "we never get there" is "why they are irrational" is nonsense. The sequence $0.3,\ 0.33,\ 0.333,\ 0.3333,\ \ldots$ approaches $1/3$ and never gets there, but $1/3$ is rational. ${}\qquad{}$ – Michael Hardy Jan 11 at 14:29
@fleablood, great question. The answer is, it doesn't. This e is not the smallest element and of course in a field you couldn't have a smallest element. As a first approximation think of a sequence tending to zero as generating an infinitesimal, e.g., $\frac{1}{n})$. Then its square will be so much smaller: $\frac{1}{n^2}$. More details can be found here: math.stackexchange.com/questions/1602977/… – Mikhail Katz Jan 12 at 16:41
@fleablood : This is not about surreal numbers, but about Robinson's hyperreal numbers. An infinitesimal $\varepsilon>0$ is smaller than all reciprocals of finite integers, and hence smaller than all positive real numbers, but it is certainly not a smallest positive hyperreal number, since $\varepsilon/2$ is smaller, and so is $\varepsilon^2$. ${}\qquad{}$ – Michael Hardy Jan 12 at 18:07

I looked over the existing answers and they don't seem to address the concern of the OP as I understood it, so I will try to provide a separate answer. The point is that the $(\epsilon,N)$-type definition of convergence is a first-order property and therefore by the transfer principle is still satisfied over the hyperreals. More specifically, a Cauchy sequence $(x_n)$ will have a natural extension defined even for infinite values of the index $n$. This extended sequence will satisfy the condition you mentioned, namely for every $\epsilon$ there is an $N$ such that, etc. If epsilon is infinitesimal then as you point out $N$ will have to be infinite typically.

Having said this, the definition in question is provably equivalent to the one mentioned by other editors (and that involves fewer quantifiers), namely if both indices are infinite then the difference $x_n-x_m$ will be infinitesimal.

-