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Homework Help: Topology, help with limit points

  1. Oct 9, 2011 #1
    Not sure if this is the correct place to post this, please move if need be.

    I am currently learning about limit points in my Topology class and am a bit confused. Going by this:

    As another example, let X = {a,b,c,d,e} with topology T = {empty set, {a}, {c,d}, {a,c,d}, {b,c,d,e}, X}. Let A = {a,b,c} then a is not a limit point of A, because the open set {a} containing a does not contain any other point of A different from a. b is a limit point of A, because the open sets {b,c,d,e} and X containing b also contain a point of A different from b. Similarly, d and e are also limit points of A. This illustration suggests that a set can have more than one limit point.

    What I do not understand is that {a,c,d} contains points different from a . Same goes for point c, {b,c,d,e} contains points other than c that are also in A but it is not a limit point? Can anyone explain this to me at all?
  2. jcsd
  3. Oct 9, 2011 #2
    Ah, I figured it out. It seems that if you find one example of there being an open set not containing the points in A it disregards it as being a limit point
  4. Oct 9, 2011 #3


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    let's look at the neighborhoods of each point:

    a: {a}, {a,c,d} , X

    b: {b,c,d,e}, X

    c: {c,d}, {a,c,d}, {b,c,d,e}, X

    d: {c,d}, {a,c,d}, {b,c,d,e}, X

    e: {b,c,d,e}, X

    now a limit point x of A has to satisfy 2 requirements:

    1) any neighborhood of x intersects A
    2) every such intersection has something besides x in it.

    all the neighborhoods of a intersect A, so the first condition is satisfied. the second condition is satisfied for {a,c,d} and X, but NOT for the neighborhood {a}. (think of a as being "away" or "isolated" from the other points in A). so a is not a limit point.

    now every neighborhood of b intersects A, so we're good on condition 1. also {b,c,d,e}∩(A-{b}) = {c}, while X∩(A-{b}) = { a,c}, so condition 2 is satisfied. thus b is a limit point of A.

    now let's look at c: {c,d}, {c,d,e}, {b,c,d,e} and X all intersect A. but {c,d}∩(A-{c}) is empty, so condition 2 fails.

    now let's look at d. all neighborhoods of d intersect A, so that's half the battle. {c,d}∩(A-{d}) = {c}, {a,c,d}∩(A-{d}) = {a,c}, {b,c,d,e}∩(A-{d}) = {b,c}, and X∩(A-{d}) = A. so d is a limit point.

    note that we could have just checked for intersection with A, since A-{d} = A (because d isn't in A).

    finally, e: as with d, A-{e} = A, so we only need to check condition 1. both neighborhoods of e intersect A, so e is a limit point of A.

    now, the motivating example of open sets, is open disks in the real euclidean plane. and we want the limit points of a set in the plane to be able to be "approximated well" by other points in the set. consider this example:

    A = {(x,y) : x^2 + y^2 < 1} U {(2,2)}.

    even though the point (2,2) is in A, we can't approximate it by other points of A, because none of them are anywhere near it, it's out there all by itself.

    and even though (1,0) ISN'T in A, any open disk around (1,0) (no matter how small) intersects with A, so we can get as "close as we like" to (1,0) with points in A.

    of course, with finite toplogies, such as the one in this problem, things don't behave like you might expect them to. you could think of the "neighborhoods" in X as being "clubs" the points of X belong to. well, one of the clubs a belongs to, just consists of a alone (how sad is that?) if you take a out of that club, there's no "sharing" with A.

    on the other hand, even those d and e aren't even IN A, every club they belong to, has a connection with A, shared with some other member (it's like they are "almost in A").

    your objection that {a,c,d} contains points of A besides a, would be valid, except that condition 2 has to hold for EVERY neighborhood of a, not just some.
  5. Oct 9, 2011 #4
    Thank you, that is a perfect explanation. I wish my book said that, I appreciate the time you took typing out your response.
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