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I have been hearing different people saying this in different contexts for quite some time but I still don't quite get it.

I know that compact operators map bounded sets to totally bounded ones, that the perturbation of a compact operator does not change the index, and that the calkin algebra is an indispensable tool in the study of operators in the sense that 'essentially something' becomes a useful notion.

But I still suspect why they are 'small'. Now Connes says they are like 'infinitesimals' in commutative function theory, which makes me even more confused. So I guess I just post this question here and hopefully I can hear some quite good explanations about the reasoning behind this intuition.

Thanks!

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    Have you ever seen in real a compact operator? I think no, because they are small.2012-07-20
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    @Norbert: what on earth does that comment even mean? Look up "integral operators" for plenty of compact operators that arise naturally from applied mathematics or physics. Or just *think* for a moment to come up with loads of examples.2012-07-20
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    If it is not too heretical to say this: occasionally one should take the pronouncements of brilliant mathematicians with a *pinch* of salt, perhaps even a whole compact operator's worth...2012-07-20
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    More generally, compactness is sometimes seen, in analysis, as "the next best thing after finiteness" -- this imprecise but useful POV is expounded on slightly in e.g. Sutherland's Introduction to Metric and Topological Spaces, and probably also somewhere on Terence Tao's blog IIRC.2012-07-20
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    @Norbert Sorry this time I do not quite get the joke :)2012-07-21
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    @YemonChoi Yes. In topology compactness is the next big thing after finiteness and this is explained by the definition of compactness I think- you can cover the set by finite open sets. But I guess the situation is not the same for compact operators. They may be still the next best thing after 'finiteness', in the sense that they exhibit 'almost finite dimension' behavior. So I guess it is okay to say they are 'small', but Connes' 'infinitesimally small' is what really confuses me.2012-07-21
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    @YemonChoi When one try to explain a joke it won't be a joke anymore. Have you ever seen the elephants in the bush tomato. I think no, because they are well hidden.2012-07-21

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It may help to think of the special case of diagonal operators, that is, elements of $\ell_\infty $ acting on $\ell_2$ by multiplication. Here compact operators correspond to sequences which tend to 0, "are infinitesimally small". This is a commutative situation, in which everything reduces to multiplication of functions. So, general compact operators can be called noncommutative infinitesimals.

A shorter explanation, but with less content: every ideal in a ring can be thought of as a collection of infinitesimally small elements, because they are one step (quotient) away from being zero.

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    I like your second explanation better actually. As for the first one, it is hard for me to see why sequences tending to $0$ are infinitesimally small. For instance, we do not usually say a function vanishing at infinity is infinitesimally small, right?2012-07-20
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    @HuiYu They are essentially small in the following sense, for example. A sequence in $\ell^\infty$ tending to $0$ may have a few big terms, but in the end, for every $\epsilon > 0$, the infinitely long tail of the sequence is smaller than $\epsilon$ whereas only a few finite terms exceed it.2012-07-20
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Finite rank operators are small (in that they squish a large space into a small one). In a Hilbert space (or more generally, a Banach space with the approximation property, which includes most familiar examples), compact operators are precisely the operator-norm limits of finite rank operators. That's what I think of when I hear that statement.

Alternatively, a compact operator from $X$ to $Y$ squishes the unit ball of $X$ (which is "big") into a compact subset of $Y$ (which, in Banach spaces, is typically "small" in that it has empty interior).

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    Yes. This is related to the fact that bounded sets are mapped to totally bounded ones by compact operators. But I guess there is something more intrinsic to the algebra of operators, not much to do with the underlying space.2012-07-20
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    ...over a Hilbert space. This is false for general Banach spaces.2012-07-20