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    [0,1] is closed, [0,1) is open/closed, so why is it that [0,infinity) is a closed set?

    I thought if you include the upper limit the set is closed, but as there is no real number for infinity how is closed?
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    (Original post by BackToMathsAgain)
    [0,1] is closed, [0,1) is open, so why is it that [0,infinity) is a closed set?

    I thought if you include the upper limit the set is closed, but as there is no real number for infinity how is closed?
    Is [0,1) not half open/half closed
    [0,infinity) is half open/half closed


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    (Original post by rayquaza17)
    Is [0,1) not half open/half closed
    [0,infinity) is half open/half closed


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    my bad on [0,1), you are correct. My notes say that [0,infinity) is closed though :s
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    (Original post by BackToMathsAgain)
    my bad on [0,1), you are correct. My notes say that [0,infinity) is closed though :s
    That's weird: my notes say it's open, but when I google it people say it's closed?

    Maybe someone with some more knowledge can help us.

    EDIT: I've checked Thomas' Calculus (twelfth edition) and on ap-3 it says that [a,infinity) is a closed set. As does Wolfram MathWorld.

    Wikipedia however (which obviously is a great source compared to the two mentioned above) says this notation is ambiguous?


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    I've just checked my notes again, and the lecturer says we could class [0,infinity) as a closed set, but the explanation is beyond the scope of the module??

    I'm in favour of this being ambiguous, but I'd like to hear what anyone else has to say??
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    In a topological sense, with the usual topology on \mathbb{R}, [0,\infty) is a closed set because the complement (-\infty, 0) is an open set (that is, for any x \in (-\infty, 0) you can find an \epsilon > 0 such that (x-\epsilon, x+\epsilon) \subset (-\infty, 0))
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    I don't think you understand what it means for a set to be closed if you cannot see why [0,∞), with its usual topology, is closed. Reread your definition, it has absolutely nothing to do with brackets. Furthermore, [0,1) is not open/closed, it is neither.
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    The way I visualize this for the real line is to imagine a sequence that converges. A set S is closed only if every sequence that converges also converges to a point in S. So in other words, [0,1] is closed because if  x_n \to x and  0 \leq x_n \leq 1 then  x \in [0,1] . Similarly,  [0,\infty) is closed because every convergent sequence will also have a limit in  [0,\infty) . Obviously  [0,1) is not closed because of  1-\frac{1}{n} .

    Edit: I hope this isn't too confusing, but a set can be open and simultaneously closed. I won't go into details, but the point is that open and closed are just names. They are also stupid names, which is why something can be open and closed simultaneously.
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    It contains all of its own limit points. So it's closed.


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    Online tutoring i the best in current situation , one can get the study while sitting in home .
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    (Original post by IrrationalNumber)
    The way I visualize this for the real line is to imagine a sequence that converges. A set S is closed only if every sequence that converges also converges to a point in S.
    Shouldn't this say "A set S is closed only if every sequence of points in S that converges also converges to a point in S"?

    By the way, it's clear that this is a very tricky topic.
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    (Original post by atsruser)
    By the way, it's clear that this is a very tricky topic.
    Wow, there really is a video like this for everything that upsets people! 😂


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