A Closed set in the Complex Field

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Discussion Overview

The discussion revolves around the characterization of a set in the complex field defined by the inequality |c – i| ≥ |c|, where c is a complex number. Participants explore the geometric interpretation of this set, its boundaries, and whether it can be classified as closed.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant asserts that the set defined by |c – i| ≥ |c| is closed and seeks to visualize it as outside a disc centered at (0,1) with a radius equal to the modulus of c.
  • Another participant suggests finding the boundary of the set as a first step in understanding its properties.
  • A different viewpoint emphasizes visualizing complex numbers as points in the plane and constructing right triangles to analyze distances from the origin.
  • One participant reiterates the initial claim about the set being closed and provides a geometric interpretation involving the perpendicular bisector of the segment from 0 to i, stating that the set consists of points closer to 0 than to i.
  • There is a correction regarding the representation of the perpendicular bisector, clarifying that it should be expressed as c = x + (1/2)i for any real x.
  • Participants engage in refining each other's statements and correcting misunderstandings about the geometric representation.

Areas of Agreement / Disagreement

Participants express differing views on the geometric interpretation and closure of the set, with no consensus reached on the characterization of the set as closed or the implications of its boundary.

Contextual Notes

Some assumptions about the geometric properties of complex numbers and the implications of the inequality are not fully explored, leaving room for further discussion.

Bachelier
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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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Hint: find the boundary first.
 
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.
 
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 |c- i|\ge |c| 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:
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.
 
Right, thanks. And thanks for trailing along behind me cleaning up my mess!
 

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