A Closed set in the Complex Field

In summary, the set |c-i| ≥ |c| with c being in ℂ is a closed set that is outside the disc centered at (0,1) with radius equal to the modulus of c, where c is a complex number with a real part of x and an imaginary part of (1/2)i. This can be visualized as the perpendicular bisector of the segment from 0 to i, where points on the side closer to 0 than to i make up the set.
  • #1
Bachelier
376
0
This is elementary but surely this set is closed

| c – i | ≥ | c | with c being in ℂ

I am trying to picture the set. Is it outside the disc centered at (0,1) with radius equal to modulus c (whatever that is) ?

Thanks
 
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  • #2
Hint: find the boundary first.
 
  • #3
Visualize the complex number as a point in the plane. Move it down one unit. (That's what subtracting i does.) Now, what points will be further from the origin when this happens? You can construct a right triangle of legs Re(c) and Im(c) and a right triangle of legs Re(c-i) and Im(c-i) and see which hypotenuse is longer.

Once you have figured out the region in question, it should be easy to show its closure.
 
  • #4
Bachelier said:
This is elementary but surely this set is closed

| c – i | ≥ | c | with c being in ℂ

I am trying to picture the set. Is it outside the disc centered at (0,1) with radius equal to modulus c (whatever that is) ?

Thanks
|c- i| is the distance from point c to i. |c| is the distance from c to 0. Saying that |c- i|= |c| (as pwsnafu suggested "find the boundary first") is the same as saying that those two distance are equal for all c. Geometrically, that is the perpendicular bisector of the segment from 0 to i, Im(c)= 1/2 or c= x+ (1/2)i for any real x. The set [itex]|c- i|\ge |c|[/itex] is set of points on ther side of that line closer to 0 than to i.

Corrected thanks to oay.
 
Last edited by a moderator:
  • #5
HallsofIvy said:
Geometrically, that is the perpendicular bisector of the segment from 0 to i, Im(c)= 1/2 or c= (x/2)i for any real x.
Not quite.

You mean: c = x + (1/2)i for any real x.
 
  • #6
Right, thanks. And thanks for trailing along behind me cleaning up my mess!
 

1. What is a closed set in the complex field?

A closed set in the complex field is a set of complex numbers that contains all of its limit points. This means that any sequence of complex numbers within the set will converge to a point within the set itself.

2. How do you determine if a set is closed in the complex field?

A set is closed in the complex field if its complement (the set of complex numbers not in the set) is open. In other words, if every point outside of the set has a neighborhood that does not intersect with the set, then the set is closed.

3. Can a set be open and closed in the complex field?

Yes, a set can be both open and closed in the complex field. This type of set is called a clopen set. An example of a clopen set is the entire complex plane, which includes all complex numbers and their limit points.

4. What is the relationship between closed sets and convergence in the complex field?

Closed sets in the complex field are closely related to the concept of convergence. If a sequence of complex numbers converges to a point within a closed set, then that point must also be within the set. This is because closed sets contain all of their limit points.

5. How are closed sets in the complex field used in mathematics?

Closed sets in the complex field are used in various branches of mathematics, such as analysis, topology, and complex analysis. They are essential in understanding the behavior of sequences and series of complex numbers, as well as in proving theorems and solving problems related to continuity and compactness.

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