# The interpretation of $0 \cdot \infty$

When I read books about measure theory (in particular Folland's Real analysis), authors note that, sometimes, they assume $0\cdot\infty = 0$. I think, it is obvious because if you sum infinitely many zero, you will obviously get zero. Then, why this is called an interpretation? Conversely, when I think about zero times $\infty$, the result should be again zero. I do not understand the interpretability of this statement. Thanks.

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What kind of books? In analysis for example, it is possible to have two sequences where that two limits tend to 0 and $\infty$ respectively, and the limit of the product of the sequences is 0, and also possible to have two sequences with limits of 0 and $\infty$ but the limit of the product is $\infty$ –  Tom Oldfield Nov 10 '12 at 15:48
The expression has no meaning because infinity is not a number. Multiplication is defined for numbers. –  Thomas Andrews Nov 10 '12 at 15:48
@ThomasAndrews, see Folland's Real analysis, p. 17. –  oeda Nov 10 '12 at 15:51
@TomOldfield There is nothing about limits in the question. –  Thomas Andrews Nov 10 '12 at 15:51
Sorry, but I can't see that book, I don't have it. How about making your question self-contained? –  Thomas Andrews Nov 10 '12 at 15:52

This is one of several situations where a common definition leads to a situation that behaves badly with respect to limits. Here the ancient notion of product, namely that $ab$ is the area of a rectangle whose sides have lengths $a$ and $b$, leads to $0\cdot\infty=0$, and this works fine when dealing with lengths and areas, as in measure theory. For another example, the simplest definition of exponentiation of natural numbers, namely that $a^b$ is the number of functions from a $b$-element set to an $a$-element set, leads to $0^0=1$, and this works fine when we deal with combinatorics of finite sets.

Once limits enter the picture, these definitions require some caution. When calculating limits, one frequently uses "facts" like, if $\lim_{x\to q}f(x)=a$ and $\lim_{x\to q}g(x)=b$, then $\lim_{x\to q}(f(x)g(x))=ab$ and $\lim_{x\to q}f(x)^{g(x)}=a^b$, and these become false with the definitions in the preceding paragraph. My own inclination is to say "OK, multiplication and exponentiation aren't continuous at these points, so you'd better apply these "facts" only at places where continuity holds." (In other words, don't use continuity at points of discontinuity.) Many people (including especially calculus students) get in the habit of just using these "facts" while ignoring the need to check any hypotheses. To prevent these people from getting wrong answers, other people (including especially the authors of calculus texts) have adopted the policy of calling things like $0\cdot\infty$ and $0^0$ indeterminate forms. This is a way of warning people loudly to avoid using continuity in these cases. That is, instead of saying $0^0=1$ but watch out when using this in limits, one says $0^0$ is undefined, so you can't evaluate it in limits.

Since people usually learn calculus early in their mathematical careers, they get indoctrinated with the "knowledge" that $0\cdot\infty$ and $0^0$ (and other such things) are undefined. As a result, when a measure theorist wants to use the natural definition, which makes $0\cdot\infty=0$, or when a combinatorialist wants to use the natural definition that makes $0^0=1$, they must point this out explicitly. The bold ones will say "the correct definition gives this"; the more timid ones, who want to remain at peace with the calculus books, will say "we adopt this convention."

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However, I wish people would note the distinction between "indeterminate form" and "undefined expression". –  Hagen von Eitzen Nov 1 '13 at 21:22

As remarked in the comments, $0\cdot \infty$ makes no sense as an ordinary product. There are situations where it is useful to agree that $0 \cdot \infty =0$: the best known situation is abstract measure theory.

But this is a convention that cannot preserve the continuity of the product, as the well-known examples of indeterminate limits show.

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For example: Suppose $\mu(X) = \infty$. We want the integral of non-negative simple functions to satisfy $\int \sum_{i=1}^n c_i\chi_{E_i} \,d\mu = \sum_{i=1}^n c_i\mu(E_i)$. In particular for the $0$ function: $0 = \int 0 \,d\mu= \int 0 \cdot \chi_X \,d\mu = 0 \cdot \int \chi_X \,d\mu= 0 \cdot \mu(X) = 0 \cdot \infty$. –  kahen Nov 10 '12 at 16:39

Books written by mathematicians never assume something like $∞·0=0$. It doesn't make sense. It's not even a mathematical catch, it's a clear truth you can see from examining how functions behave. $Let me give you an example.$f(x)=x$gets closer to infinity as$x$gets closer to infinity, that much is obvious. It's also clear that,$f(x)·0=x·0=0$, and that just stays zero no matter how$x$behaves. However,$\frac{1}{x}$also approches$0$as$x→∞$. But for example,$\frac{1}{\sqrt{x}}·\frac{1}{x}=\frac{1}{\sqrt{x}}$which also approaches$0$as$x→∞$. But$x·\frac{1}{x} = 1$which actually stays$1$no matter how x behaves. In crude and inaccurate terms, you can say$∞·0$equals different things depending on how 'fast' or how 'powerful' those$0$and$∞$are. That said, in some areas, a variant of that equation can be true. This is usually where we're speaking about values that have a definite, but still infinite size. For example, if$A$and$B$are sets,$A$is an infinite set, and$B$is a set that has$0$items (an empty set, which has size$0$), then$A×B$is also a set that has$0$elements. This is what similar to the idea that no matter how many zeroes you have, the thing still remains zero. Systems such as surreal numbers also define values of infinite size. In fact, there are values that are infinitely more infinite than an infinite other infinite infinities, but in all of those, it is still the case that$ω·0=0$, where ω is said infinite number. However, the notation$∞$refers to a specific kind of unbounded infinity. It's not a value in the normal sense. Nor is it a number you can manipulate. - "Books written by mathematicians never assume something like$\infty \cdot 0 = 0$" Both Rudin's Real and Complex Analysis and Halmos' Measure Theory make exactly that definition. It is a completely standard convention in measure theory. This is because the integral of the constant zero function on a set of infinite measure is zero, and the integral of the function$f(x) = \infty$on a set of measure zero is also zero, so the multiplication formula works correctly for computing the integral when we set$\infty \cdot 0 = 0 \cdot \infty = 0$. – Carl Mummert Nov 10 '12 at 16:52 You can manipulate$\infty$in the extended real line. But still for that number multiplication by zero is indefinite. – Anixx Nov 10 '12 at 16:58 @CarlMummert that's true. The definition and use of the symbol ∞ is entirely dependent on the field and the context. The statement that$∞·0=0\$ makes no sense, itself makes no sense. I was a tad confused in my further and overlooked that fact. –  GreĝRos Nov 10 '12 at 17:11