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Questions on Bolzano-Weirestrass, Discreteness

First, i would like to show that Z subset of R is discrete. I think i've done this, but would appreciate feedback. This is what I did: consider any a in Z. Let e=0.5. Then B_(0.5)(a) n Z = {a}. Therefore, Z is discrete. (Q: is it not allowed to choose e>1, since that would mean that the intersection of the ball and Z has more than one integer? Is any e<=1 fair game?)

Now, I want to show that f:s-smilie:-->R is continuous if S is discrete, but I don't know how to relate epsilon-balls to continuity... Since S is discrete, there exists e>0 s.t. B_e(x) n S = {x} for all x in S. What do i do from here?

Finally, every closed, bounded, discrete set is finite. Why is each of these conditions necessary?
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Jonny W
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J.F.N
First, i would like to show that Z subset of R is discrete. I think i've done this, but would appreciate feedback. This is what I did: consider any a in Z. Let e=0.5. Then B_(0.5)(a) n Z = {a}. Therefore, Z is discrete. (Q: is it not allowed to choose e>1, since that would mean that the intersection of the ball and Z has more than one integer? Is any e<=1 fair game?)

You need to find, for each point in Z, an e > 0 that satisfies the property. For each point in Z, e = 1/2 is indeed suitable.
J.F.N

Now, I want to show that f:s-smilie:-->R is continuous if S is discrete, but I don't know how to relate epsilon-balls to continuity... Since S is discrete, there exists e>0 s.t. B_e(x) n S = {x} for all x in S. What do i do from here?

For any y in S with |y - x| < e we have y = x and |f(y) - f(x)| = 0. (Perhaps you should say "d" rather than "e", since the latter usually refers to the thing that |f(y) - f(x)| must be less than.)
J.F.N

Finally, every closed, bounded, discrete set is finite. Why is each of these conditions necessary?

"Closed and bounded" is not enough
Eg, S = [0, 1]

"Closed and discrete" is not enough
Eg, S = N

"Bounded and discrete" is not enough
Eg, S = {1/n : n in N}

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Let S be closed, bounded and discrete. Suppose that S is infinite. Let (x_n) be a sequence in S with all its terms distinct. Since S is bounded, there is a subsequnce (y_n) of (x_n) and an L in R such that y_n -> L. Since S is closed, L is in S. Since S is discrete, there is an epsilon > 0 such that |x - L| >= epsilon for all x in S \ {L}. Then |y_n - L| >= epsilon for all n with at most one exception. (The possible exception is an n such that y_n = L.) So y_n does not tend to L. Contradiction.

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