What is the definition of infinity from the point of real infinity?
How is this different from the concept of infinity by the ZFC axiom?
Consider any diverging sum for example $ \sum_{1}^{\infty }{n^{2}} $.? Does this expression even make sense?

Another question is that what do we mean by a bigger infinity? We know cardianlity of power set is strictly greater than cardinality of the set, so the cardinality of reals is greater than natural numbers, so does the ZFC axiom discriminate between these two types of infinity, even though Cantor proved that reals contain 'more' element than natural numbers?


Saying $\mathbb{Z} = \{ \dots,-2,-1,0,1,2,\dots \}$ (the set of integers) is an infinite set is using a different (but not totally unrelated) notion of infinity than say $\displaystyle\sum\limits_{n=1}^\infty n^2=\infty$.

The main difference between the two uses of infinity is that the first has to do with cardinality (essentially size) and the second has more to do with ordering.

In set theory, one says that two sets have the same cardinality if there is a bijection between them (an invertible map). So for example: $X=\{a,b,c\}$ and $Y=\{1,2,3\}$ both have that same cardinality.

One can prove that the real numbers, $\mathbb{R}$, can be put into bijective correspondence with the power set of the natural numbers: $\mathcal{P}(\mathbb{N})$ (i.e. the set of all subsets of $\mathbb{N}=\{0,1,2,\dots\}$). So $\mathbb{R}$ and $\mathcal{P}(\mathbb{N})$ have the same cardinality (we write: $|\mathbb{R}|=|\mathcal{P}(\mathbb{N})|$).

Cantor proved (it's not too hard to do so) that $|X|<|\mathcal{P}(X)|$ (a set has smaller cardinality than it's power set). This means that while there is a one-to-one map from $X$ to its power set, there isn't a bijection between the two sets. So a power set is always bigger than the set its built from.

This means that even though $\mathbb{N}$ and $\mathbb{R}$ are both infinite sets, $\mathbb{R}$ is (in this sense) a bigger infinite set.

So $\mathbb{R}$ is a big infinite set.

On the other hand, $\mathbb{R}$ is a totally ordered set (for each $a,b\in\mathbb{R}$ we have either $a<b$, $b<a$ or $a=b$). Saying $\displaystyle\sum\limits_{n=1}^\infty n^2=\infty$ means that as we try to sum this series the sum exceeds (is bigger than) all real numbers. This time bigger is used in the ordering sense (ordering not size).

Finally, can we make sense out of $\displaystyle\sum\limits_{n=1}^\infty n^2$? Sure.

One option is to campactify the real numbers by tacking on $\pm \infty$. So instead of working with $\mathbb{R}=(-\infty,\infty)$ we could work with $\hat{\mathbb{R}}=\mathbb{R}\cup\{\pm \infty\} = [-\infty,\infty]$. Although the set $\hat{\mathbb{R}}$ no longer has all of the same nice (algebraic) properties as $\mathbb{R}$, we now have that $\displaystyle\sum\limits_{n=1}^\infty n^2=\infty$ where $\infty$ can be treated like any other number.

Another option for making sense out of some divergent series is to treat them formally (i.e. don't worry about convergence). Euler did a lot with divergent series in this way. Generating functions arise in this context.

I hope that clears up some of the confusion.

Edit: The notation "$\sum\limits_{n=0}^\infty n^2$" means to compute all finite sums $\sum\limits_{n=0}^N n^2$ and then see if these sums have a limit as $N$ get arbitrarily large (i.e. "goes to infinity").

So again the "$\infty$" here again refers to the ordering. Summing up to "$\infty$" means sum up as far as possible. Since we can only add up finite sums, we take a limit to see what "happens at infinity".

As for treating the series formally, I should be more careful than what I originally wrote. You don't want to think about "summing up" the series at all. Here we start to use symbols being summed up as placeholders. This allows series manipulations to capture combinatorial phenomena.

For example: Think of $x \cdot x = x^2$ as capturing $1+1=2$. Likewise, $x^2 \cdot x^6 = x^8$ as capturing $2+6=8$.

If we want to figure out how many partitions of 5 (i.e. ways of adding positive integers up to get 5) there are, we can just list them out... $5=4+1=3+1+1=3+2=2+2+1=2+1+1+1=1+1+1+1+1$. So there are 7 partitions of 5.

On the other hand, we could make series do this for us. $1+x+x^2+x^3+x^4+\cdots$ captures sums of 1's and $1+x^2+x^4+x^6+\cdots$ captures sums of 2's and so on. If we multiply $(1+x+x^2+\cdots)(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)\cdots$ out formally we get $1+x+2x^2+3x^3+5x^4+7x^5+\cdots$.

The multiplication of these series captures of the number of partitions as the coefficient of $x^?$. This works because multiplying powers of $x$'s turns into addition of exponents. Thus the number of ways to add up to 5 is the same as the number of ways to have $x^5$ show up after these series are multiplied together.

So suppose that $p(n)$ is the number of partitions of $n$ (for example: $p(5)=7$). Then we have: $$\sum\limits_{n=0}^\infty p(n)x^n = (1+x+x^2+\cdots)(1+x^2+x^4+\cdots)\cdots = \frac{1}{1-x} \cdot \frac{1}{1-x^2} \cdots = \prod\limits_{n=1}^\infty \frac{1}{1-x^n}$$

With care these infinite sums and products can be manipulated to prove all sorts of things. But we never assign an actual value to $x$. It's just a place holder. So we're not really "summing up" anything.

Also, even though these are "infinite" sums/products, any computation we do with them is really finite. For example, to compute $p(5)$ (the coefficient of $x^5$), we only need to involve finitely many of the series and only finitely many terms of the series that actually matter, so in the end we have a finite computation.

  • $\begingroup$ Ok actually i wanted to completely understand the abstractness of summation, first i was thinking what do you mean when you sum from 1 to infinity that is how many times do you do addition, then i thought since infinity is not a part of N, what we mean is that we are doing the addition for a 'large N' and since the series is convergent, we are given the freedom to 'notationally' represent this concept by using the summation from 1 to infinity. Simply speaking, my next question is what do you mean, formally when you sum a divergent series upto infinity in case of traditional real analysis or N $\endgroup$ – Varun Jan 5 '15 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.