{m+1/n|n is a natural} given m a natural has M as ONLY limit point

[faradayslaw]

Homework Statement

I wish to show {m+1/n|n is a natural} given m a natural has M as ONLY limit point

Homework Equations

Rationals dense in reals

The Attempt at a Solution

Let M = {m+1/n|n is a natural}.
I can easily show, by archemidean principle of Reals that m is a limit point of M. I need to show no other real y is a limit point.

I try to define a set of points K= {j| j is in M and d(j,y) <= d(k,y) for all k in M).

Essentially if I have that K is nonempty, then since 0<d(y,k) for all k in K, there is q in Q s.t. 0<q<d(y,k) for all k in K, and so a neighborhood of radius q/2 and center at y contains no points of M -> y is NOT a limit point of M.

THe problem I realized is that m is a number not in M, and so the above situation could be applied to m replacing y in the above, and that I really need to show that the d(j,y) can not be taken arbitrarily small i.e. that K is non empty for y =/= m. I am unsure on how to do this and any help is appreciated.

Thanks,

 

[mighty2000]

Scientist

Dear Scientist,

Thank you for sharing your thoughts and approach to this problem. Your use of the Archimedean principle is a good start in showing that m is a limit point of M. To show that no other real number is a limit point, you can use a proof by contradiction.

Assume there exists another real number y that is a limit point of M, where y ≠ m. This means that for any ε > 0, there exists a point j in M such that 0 < |j - y| < ε. However, since M = {m+1/n|n is a natural}, we can choose a natural number n such that |m+1/n - y| < ε. This implies that |m - y| < ε, which contradicts our assumption that y ≠ m.

Therefore, we have shown that m is the only limit point of M, and we can conclude that M = {m+1/n|n is a natural} has M as its only limit point.

I hope this helps! Keep up the good work in your scientific pursuits.Your colleague