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Homework Help: Determining open and connected sets.

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Are the following regions in the plane (1) open (2) connected and (3) domains?
    a. the real numbers;
    b. the first quadrant including its boundary;
    c. the first quadrant excluding its boundary;
    d. the complement of the unit circle;
    f. C \ Z = {z ∈ C : z [itex]\notin[/itex]Z}.

    3. The attempt at a solution

    a) Open, connected, domain - because there is no boundary, all points on the plane will be interior points of the set of real numbers, and it will clearly be a connected set. Hence a domain.

    b) It would be a closed set because if you picked the boundary point, an ε-neighbourhood would have a region outside this boundary, and hence a point not in the set. However, the set would be connected, can not be a domain.

    c) A closed set, and connected because you should be able to get from one point to another point in the quadrant as long as it's within the boundary. Not a domain

    d) Another closed set, and it is connected as long as it's within the boundary again. Not a domain again because it's not open and connected.

    f) I have no idea what (f) is referring to

    Are these correct? I'm not entirely sure of these answers or their reasonings. Also, I don't understand part f.
  2. jcsd
  3. Mar 11, 2012 #2
    For part f is the notation supposed to read: [itex]\mathbb{C} \backslash \mathbb{Z} = \{ z \in \mathbb{C}: z \notin \mathbb{Z}\}[/itex]? Then that set is simply the elements in the complex plane excluding the integers. So, [itex]\mathbb{C} \backslash \{... -2, -1, 0, 1, 2, ...\}[/itex]

    Can you give us your definitions/theorems of being closed? Complements of open sets, containing all of its limit points, etc. I think then we can help you understand why they are open/closed. I noticed your answers for the open and closed parts weren't all correct and I can explain why if I have the definitions/theorems you have at your disposal up to this point.
  4. Mar 11, 2012 #3

    Let S [itex]\subset[/itex] ℂ

    An interior point of S is a point z[itex]_{0}[/itex] [itex]\in[/itex] S such that there is an [itex]\epsilon[/itex]-neighbourhood of z[itex]_{0}[/itex] contained in S.

    A boundary point of S is a point z1 such that every [itex]\epsilon[/itex]-neighbour hood of z1 contains points in S and not in S.

    S is open if all its elements are interior points.
    S is closed if it contains all its boundary points.

    S is connected if two points in it can be connected by a continuous curve or line in S.

    An open and connected set is called domain.

    Those were the definitions we were given more or less. I don't have a definition for complement of open set. Although I just visualised/drew out the boundary/circle and then took the complement of it and tried to apply what I already knew.
  5. Mar 11, 2012 #4
    Yes, that's exactly the notation.

    O.K. So you're saying ℂ\ Z means the elements in the complex plane excluding the integers? And the r.h.s of that expression was just the same thing?
  6. Mar 11, 2012 #5
    Okay, that's good, sometimes in math you start a subject from different definitions for concepts and I just wanted to make sure we were on the same page.

    Like I said in my last post, you seem to get mixed up on open and closed. Since your definition of closed involves the boundary points of the set, can you show what the boundary is for parts c and d? You mentioned they were both closed but gave no reason for why. Try also giving the boundary of part f and see what you come up with.

    For part d, I'm not sure what you are describing for being connected. I'm just trying to visualize your thinking process.

    For your second reply, yep, that's exactly what it means!
  7. Mar 11, 2012 #6
    Well for part (c) - it's excluding its boundary points, so that would mean < #some value.

    Therefore no points on the boundary would be a part of the set. Usually we draw a continuous closed line if it's < #some value, as opposed to ≤ #some value where the boundary points can also be included.

    Given that the boundary line is not included. If we were to take a point close to the boundary line and have an ε-neighbourhood, this would start to include points outside the boundary as well as inside it. So this would mean the ε-neighbour hood has points which are not all interior points and so it cannot be open.

    This is what I'm kind of visualising:

    http://img710.imageshack.us/img710/8077/89967955.jpg [Broken]

    For part d. I drew the unit circle and shaded everything outside the circle, and considered no points on the boundary line of the circle because the complement of it should only have points not a part of the circle, and the points on the boundary are points of the circle. Hence I considered it closed because any ε-neighbourhood close to that boundary line in the complement region could theoretically have points coming into the unit circle even though the boundary is not included, so it cannot be an open set because then some points would not be in the set.

    I am trying part f now.
    Last edited by a moderator: May 5, 2017
  8. Mar 11, 2012 #7
    I think for part (f) - this would effectively just be the vertical line (y-axis) for the complex plane, with an exclusion at the origin (0,0) So this would be a closed, not connected set, and hence not a domain.
  9. Mar 11, 2012 #8
    Okay, I think you are messing up your definitions a little bit. Let me try to help clarify:

    To be an interior point, there's EXISTS an [itex]\epsilon > 0[/itex] such that that epsilon neighborhood is contained in the set S. So for part c, consider the point (1,1). You can put an open disc (circle without its boundary) around the point (1,1) with radius [itex]\frac{1}{2}[/itex] and all of those points are contained within the first quadrant. Take a closer point, say [itex](\frac{1}{4},\frac{1}{4})[/itex]. A radius of [itex]\frac{1}{8}[/itex] around that point would be completely contained in the first quadrant. Take an arbitrarily close point to the boundary of the first quadrant, say [itex](\frac{1}{n},\frac{1}{n})[/itex]. Then the open disc with radius [itex]\frac{1}{2n}[/itex] around that point will be contained in the set. So, all points are interior points and by your definition the set is open! Hopefully that makes sense.

    Alternatively, it can't be closed because it excludes its boundary points. Read the definition for closed again!

    Okay, for part d they are asking for the complement of the unit circle, not disc. So every point inside of the circle is also in the set. That should change the answer you gave for it.

    For part f the set is excluding the points [itex]... -2+0i, -1+0i, 0+0i, 1+0i, 2+0i, ...[/itex]. Your description of the set isn't correct. :( Do you understand what the set is describing now? It's all the points in the complex plane, minus the integer values on the real axis.
  10. Mar 11, 2012 #9
    Hmm, yea it's getting somewhat clearer now. I'm thinking I may have confused my including and excluding around. When they say including the boundary points. They mean ≤ right? So a point on the boundary line can also be included?

    Although if you took that point (1,1) for e.g. and had a radius of 5. Then you'd get some points well outside the first quadrant, so how would it be open then?
  11. Mar 11, 2012 #10
    Edit: NVM, to be an epsilon neighbourhood it must be contained within that set, so it can't possibly be radius 5 to begin with.
  12. Mar 11, 2012 #11
    Yep, correct! And excluding would be <.

    Nope, you're thinking the opposite way. The definition you gave for interior point was An interior point of S is a point z0 ∈ S such that there is an ϵ-neighbourhood of z0 contained in S.

    The is is not very precise. What is means is that there EXISTS. So, if you find just one epsilon such that the neighborhood is contained within the set, you are done; it's an interior point. It doesn't matter if there are bigger values that don't work, as long as you found one that works you are good.

    Compare this to the definition of boundary point. A boundary point of S is a point z1 such that every ϵ-neighbour hood of z1 contains points in S and not in S. The key word here is EVERY. If you choose epsilon equal to 3 it will work, epsilon equal to 10 it will work, epsilon equal to 1000 it will work, etc. Every choice of epsilon has to work.

    To be an interior point, there has to exist at least one epsilon. For the boundary, every epsilon. Hopefully that clears it up!
  13. Mar 11, 2012 #12
    Yes, it clears up the definitions for me. I think I'll just need a lot more practice with it to solidify it. Thanks so much for the help, it really means a lot!
  14. Mar 11, 2012 #13
    No problem! It's incredibly hard to describe topological concepts over the internet, it's much easier in person. Visualization helps a lot which is lacking here. Good luck with the questions!
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