Cardinalic flaw of Riemann integral

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I have learned that integral is the Riemann sum of infinite rectangle, that:
Ʃ^{n=1}_{∞}f(xi)Δxi = ∫^{b}_{a}f(x)dx
However, I think that (a,b) is the continuous interval, so the number of rectangle should be c instead of \aleph0 (cardinality of natural number N).
So I wonder whether there are some problem that this definition is not valid anymore.
 
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How so? The oo you're using is the countable infinity. An uncountable sum will

necessarily diverge , unless only countably-many are non-zero. Still, good

question.

Edit: after reading SteveL's comment, I guess I should be more precise:

The limit in the sum you describe is a limit as you approach countable infinity;

so you are selecting one point x_i* in each subinterval , and , as N-->oo (countable

infinity) there is a bijection between the number of rectangles and the x_i* you choose.

Since the x_i* are indexed by countable infinity, so are the rectangles.
 
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Each Δxi is a continuum - there is no contradiction.
 
pyfgcr said:
I have learned that integral is the Riemann sum of infinite rectangle, that:
Ʃ^{n=1}_{∞}f(xi)Δxi = ∫^{b}_{a}f(x)dx

I'm a little confused about this definition. Typically the Riemann integral is the limit of Riemann sums, each one of which is a finite sum over a partition of the interval. Each partition is a finite set of subintervals.

There is no infinite sum such as you've notated. Is this a definition you saw in class or in a book?
 
Thanks for explanation, I have understood.
And I mean it's the limit of finite sum, but I am a bit lazy so I remove the limit part for convenience.
 
pyfgcr said:
I have learned that integral is the Riemann sum of infinite rectangle,
No, it isn't. It is a limit of Riemann sums, each of which involves a finite sum. That is not "the Riemann sum of infinite rectangles" which is not defined.
that:
Ʃ^{n=1}_{∞}f(xi)Δxi = ∫^{b}_{a}f(x)dx
However, I think that (a,b) is the continuous interval, so the number of rectangle should be c instead of \aleph0 (cardinality of natural number N).
So I wonder whether there are some problem that this definition is not valid anymore.
It should be no surprise that your mistaken definition is not valid.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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