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How intuitive is the idea of completeness?

  1. May 3, 2013 #1
    Having read the formal definition of completeness in a metric spaces intuitively i feel asthough in Rn any subspace that every two points can be joined by a line is connected. Is this correct? I ask simply because often my simply intuitive understanding of things gets me into trouble. For sample in R2 any continuous function as a subset is connected?
     
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  3. May 4, 2013 #2

    Dick

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    You are going to have to be a lot more specific about what you are asking. Completeness and connectedness aren't that closely related. And points in R^2 aren't continuous functions. What are you getting at?
     
  4. May 4, 2013 #3
    This is rather embarrassing but I didn't mean to type completeness i meant connected.
    So what i meant to say was...

    Having read the formal definition of CONNECTEDNESS in a metric spaces intuitively i feel asthough in Rn any subspace that every two points can be joined by a line is connected. Is this correct? I ask simply because often my simply intuitive understanding of things gets me into trouble. For sample in R2 any continuous function as a subset is connected?
     
    Last edited: May 4, 2013
  5. May 4, 2013 #4

    micromass

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    The notion of path connected says that ##A\subseteq \mathbb{R}^n## is path connected if for any ##x,y\in A##, there exists a continuous function ##\gamma:[0,1]\rightarrow A## such that ##\gamma(0)=x## and ##\gamma(1)=y##.

    The notion of connected says that ##A\subseteq \mathbb{R}^n## is connected if every continuous function ##T:A\rightarrow \{0,1\}## is constant (thus either ##T(x)=0## for all ##x\in A## or ##T(x)=1## for all ##x\in A##).

    Your quote is correct in the sense that any path connected space is connected. The converse is not true though.

    But all of this doesn't really have much to do with completeness...
     
  6. May 4, 2013 #5

    micromass

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    I'm really not getting what you mean with "a continuous function as a subset".
     
  7. May 4, 2013 #6

    WannabeNewton

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    I'm having trouble parsing your sentence. If you are asking "can every subspace of ##\mathbb{R}^{n}## be connected" then unfortunately no. Take the subspace ##\mathbb{R}\setminus \left \{ 0 \right \}##. This is not connected (hence not path connected). Loosely speaking, you would have to jump past the hole when going from one path component to the other but this is not continuous (path implicitly includes continuous). In ##\mathbb{R}^{2}## if you take the subspace obtained by removing an entire line then you will run into the same problem and in ##\mathbb{R}^{3}## you can remove a plane and again get the same problem and so on and so on. This all ties in to the more general concept of homology but that is a song best saved for another day :smile:

    On the other hand, if you are asking "if a subspace of ##\mathbb{R}^{n}## is such that every pair of points in the subspace can be connected by a line" then yes the subspace is connected because it would be path connected (lines in ##\mathbb{R}^{n}## are paths of course).

    I missed this the first time reading your post so I'll address it now. Again it is not clear to me exactly what you are saying but if ##A\subseteq \mathbb{R}^{n}## is connected and ##f:\mathbb{R}^{n}\rightarrow \mathbb{R}^{m}## is continuous then ##f(A)\subseteq \mathbb{R}^{m}## is also connected (this in fact holds for any two topological spaces, not just euclidean space).
     
    Last edited: May 4, 2013
  8. May 4, 2013 #7
    The idea of something being path connected is new to me and that is essentially what i was asking. Now that i know more I guess I was asking if a path connected space was always connected. As you guys have clarified the answer is yes. Seems obvious now but thanks for clarifying that.

    What I mean by a continuous function as a subset is something like A contained in R2 as
    A={(x,y):y=sinx}. So is that connected? Seems to me that it would be path connected.

    Also micromass says that path connected spaces are connected but not necasarily the inverse. Are there any examples in Rn of a connected but not path connected space? I know one cannot think of all metric space concepts with real numbers but when possible it is always nice.
     
  9. May 4, 2013 #8
    I want to add that it seems obvious that A(as above) is path connected but I found a theorem that states a subset of R^2 is connected if and only if it is polygonally connected. I don't think A is polygonally connected which suggests that it can't be path connected. Which explains a small part of my confusion.
     
  10. May 4, 2013 #9

    WannabeNewton

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    These are all beautiful questions! As for your first question, yes ##A## is path connected. Let ##p = (x_0,\sin x_0),q = (x_1,\sin x_1)\in A##. Note that ##f:[x_0,x_1]\rightarrow \mathbb{R}^{2},x \mapsto (x,\sin x)## is continuous on the given domain because the component functions are continuous on that domain hence by the characteristic property of the product topology, ##f## is continuous as well thus making it a path from ##p## to ##q## in ##A##.

    As for your second question, the most common and also most elegant example of a space that is connected but not path connected is the topologists' sine curve: http://en.wikipedia.org/wiki/Topologist's_sine_curve

    Try to prove that it is connected but not path connected (and, if you've seen the following concept before, show it is not locally connected as well). The easiest way to show it is not path connected is using compactness but if you haven't seen compactness yet then you could finding a different way to prove it if you want.
     
  11. May 4, 2013 #10

    micromass

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    That theorem probably also says that ##A## must be open. If ##A## is an open subset of ##\mathbb{R}^n##, then the notions of connected, pathwise connected and polygonally connected are equivalent. If ##A## is not open, then there are counterexamples, as your example shows.
     
  12. May 4, 2013 #11
    Having had a second look it does say A must be open. Well spotted micomass.

    The example of the topologists sine curve is perfect, thanks. I have encountered the idea of compactness but the relationship between compactness and connected isn't obvious to me. It seems plausible that all compact spaces are connected but I'm not certain.

    I know that the union of two disjoint connected sets is connected. Singletons are trivially connected and using a very similiar arguement to the one wannabenewton used above we can prove that sin(1/x) is connected and therefore sin(1/x) union the origin is connected.
     
    Last edited: May 4, 2013
  13. May 4, 2013 #12
    The lemma that the union of two disjoint connected sets is connected doesn't instinctively make sense. The more I think about it the less sense it makes. The intervals (0,1) and (3,4) are individually path connected and therefore connected but the union of the two isn't connected.
     
    Last edited: May 4, 2013
  14. May 4, 2013 #13

    WannabeNewton

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    Sadly no this isn't true. Consider as a very trivial counter-example any non-singleton set ##X## with the discrete topology. Any finite subset ##A\subseteq X## will be compact but it won't be connected because every subset of ##A## will be clopen.

    I know you corrected this statement in your subsequent post and it is indeed not true that the union of two disjoint sets is connected. In fact what we can say is that if ##X## is a space and ##\left \{ A_{\alpha} \right \}_{\alpha\in I}## is a collection of connected subspaces of ##X## with a point in common then ##\bigcup_{\alpha\in I}A_{\alpha}## is connected.

    By the way, what book are you using?
     
  15. May 4, 2013 #14

    micromass

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    Where did you see this lemma?
     
  16. May 4, 2013 #15
    I'm just using course notes and while I would like to blame them for my mistakes I can't. Simply because they are correct but the notes I had copied down aren't. You and the course notes disagree with my notes so I clearly just copied them down wrong. The notes confirm that two connected sets must have a point in common.

    I'm gonna try and have a look at the connectedness of the topologists sine curve again and then get back to you.

    Thanks for bearing with me through my stumbles.
     
  17. May 4, 2013 #16

    WannabeNewton

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    Good luck gottfriend! It is a very instructive problem.
     
  18. May 4, 2013 #17
    I think this problem has defeated me.Atleast for now, The best I can do is prove that the closure of {sin(1/x)} is {sin(1/x)}U{(0,y) for real y}. Then prove that {sin(1/x)} is path connected and therefore connected. Then show that the closure of a connected set is connected. Therefore {sin(1/x)}U{(0,y) for real y} is connected.

    I tried to use compactness but I really don't see the relationship between compactness and completeness.
     
  19. May 4, 2013 #18

    micromass

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    This is all fine. What you used here is that if a set ##A## is connected, then ##\overline{A}## is connected too. We can refine this result as follows: if ##A## is connected and if ##A\subseteq B\subseteq \overline{A}##, then ##B## is connected. Using this result will give you the theorem.

    I think you mean connectedness. You won't need compactness to prove that the topological sine curve is connected. However, you will need compactness to prove that it's not path connected.

    And, for the record, there is a relationship between compactness and completeness. The relationship is that any compact metric space is complete. But I'm sure you didn't mean to say completeness.
     
  20. May 4, 2013 #19
    My brain is fried and all your assumptions are correct.

    Thanks for your help
     
  21. May 4, 2013 #20

    WannabeNewton

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    Try showing connectedness first, it is much easier than showing failure of path connectedness for the topologists' sine curve. If you try contradiction to show connectedness then it will be straightforward.
     
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