What is a meaning of homotopic in Cauchy's theorem for the sets?

[m_s_a]

Homework Statement

three quistions':
1) What is a meaning of (homotopic) in Cauchy's theorem for the sets'?
and in sets what is t,s where H(t,s)

2)
(-1)^1/2*(-1)^1/2 = (-1*-1)^1/2=1^1/2=1=/=i^2




3)
In Integral functions' by Reiman Sum
is delta x
It handles any inclination at the point
Concluded that by diminishing its value to zero

Homework Equations

The Attempt at a Solution

2) (-1)^1/2*(-1)^1/2 = i(1)^1/2*i(1)^1/2=i*i=-1=/=1
But the former right way
The result is wrong
logic Sports saysThe right does not lead to an error

 

[HallsofIvy]

Homework Statement

three quistions':
1) What is a meaning of (homotopic) in Cauchy's theorem for the sets'?

Two paths in a topological space (or the functions defining those paths) are said to be "homotopic" if one can be "continuously deformed" into the other. More specifically, if f:R-> X is a function from the real numbers to a space X (so that f defines a one-dimensional path in X) and g:R->X is another, then f and g are homotopic if and only if there exist a function $\Phi(t, u)$ where $0\le t\le 1$, u is real number, continuous in both t and u, and such that $\Phi(0, u)= f(u)$, $\Phi(1, u)= g(u)$.

and in sets what is t,s where H(t,s)

I don't recognize this. What is H? What area of "sets" do you mean? Since you mention "homotopic" could this be a homotopy group?

2)
(-1)^1/2*(-1)^1/2 = (-1*-1)^1/2=1^1/2=1=/=i^2

No, a^xb^x= (ab)^x does not, in general, hold for complex numbers.

3)
In Integral functions' by Reiman Sum
is delta x
It handles any inclination at the point

"Inclination"? I think that's a mis-translation. delta x measures the slight change in x on which you are basing your Riemann Sum.

Concluded that by diminishing its value to zero

It's a bit more complicated than that. You must let delta x go to 0 while, at the same time letting the number of terms in the sum go to infinity. In any case, I see no question here.

Homework Equations

The Attempt at a Solution

2) (-1)^1/2*(-1)^1/2 = i(1)^1/2*i(1)^1/2=i*i=-1=/=1
But the former right way
The result is wrong
logic Sports saysThe right does not lead to an error

"logic" doesn't work if you start from incorrect premises. As I said above, a^xb^x is not generally true for complex numbers. It is the "former" that is wrong. The definition of "1/2 power" or square root, is that (a^1/2)(a^1/2)= (a^1/2)²= a for any complex number a. (-1)^1/2(-1)^1/2= -1 is correct.

 

[m_s_a]

Thanks you the abundant thanks