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Showing an interval is equal to a set with modulus watch

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    Sorry for the not very well thought out title, but it demonstrates how destroyed my brain is at the moment.

    I'm trying to prove that for real numbers d > 0 and k that

    (k - d, k + d) = {x | |x - k| < d}

    This is intuitively/diagrammatrically quite obvious I think. However, I just can't begin to prove it properly. Should I be showing they're subsets of each other? Or is there a better way?

    What I did come up with was to say that let S = (k - d, k + d) then inf(S) = k - d and sup(S) = k + d, which is also true for the RHS of the thing we're trying to prove, and since the sets are both continuous, they must be equal. That's quite weak though because "since the sets are both continuous" isn't meaningful without more explanation (which I don't know how to give).
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    (Original post by Swayum)
    Sorry for the not very well thought out title, but it demonstrates how destroyed my brain is at the moment.

    I'm trying to prove that for real numbers d > 0 and k that

    (k - d, k + d) = {x | |x - k| < d}

    This is intuitively/diagrammatrically quite obvious I think. However, I just can't begin to prove it properly. Should I be showing they're subsets of each other? Or is there a better way?

    What I did come up with was to say that let S = (k - d, k + d) then inf(S) = k - d and sup(S) = k + d, which is also true for the RHS of the thing we're trying to prove, and since the sets are both continuous, they must be equal. That's quite weak though because "since the sets are both continuous" isn't meaningful without more explanation (which I don't know how to give).
    Couldn't you just use logic like the following?:

    {x | |x - k| < d} = {x | -d < x - k < d} = {x | -d + k < x < d + k} = (k - d,k + d). Or do you want to prove it in a more fancy way?? I think you can do this since it's just using the definition of absolute value and simple inequality re-arranging! If you want to go for the subset idea, why don't you go via proof by contradiction (both ways), eg assume set inclusions don't hold both ways, and prove by contradiction that they must then lie inside each other. Let r={x | |x - k| < d} and let y be in r but not in (k-d,k+d), then just show that y must be in (k-d,k+d) so all elements of r must lie in (k-d,k+d), just do a similar argument for the reverse and it should prove both sets are equal no?
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    (Original post by Swayum)
    I just can't begin to prove it properly. Should I be showing they're subsets of each other? Or is there a better way?
    I'd go the subset approach I think. Suppose x\in (k-d, k+d). Then k-d < x < k+d. So |k-x| < d, so ... (and vice versa).

    What I did come up with was to say that let S = (k - d, k + d) then inf(S) = k - d and sup(S) = k + d, which is also true for the RHS of the thing we're trying to prove, and since the sets are both continuous, they must be equal. That's quite weak though because "since the sets are both continuous" isn't meaningful without more explanation (which I don't know how to give).
    I don't believe this is the right approach!
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    (Original post by DFranklin)
    I'd go the subset approach I think. Suppose x\in (k-d, k+d). Then k-d < x < k+d. So |k-x| < d, so ... (and vice versa).
    I think this is probably the easiest way and I'm pretty retarded for not having come up with it. Thanks.
 
 
 
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