I would truly appriciate, if anyone could check whether my thinking is wrong or not, and if I made any mistake point it out and show how to correct it:

In metric space $ (K,d) $ in which we have $ a_n \in K $ such that $d(a_n,a_l) \ge 1/3 $ for any $l \neq n$, I try to prove that set given as $N= {\{a_1, a_2,...}\} $ is closed on $K$ and that $K$ can't be compact.

So with 1'st one I am not sure if it's sufficient to say that if we take any $x \in K \setminus N $ we can take $r=d(x,N)$ and pick open ball around such $x$, with radius $r/2$. Than it can't intersect with $N$, it contains $x$, and it's open set in given metric induced topology.

It seems too simple, but maybe I am just nitpicking.

With 2'nd one I went like: if such space were compact, than any of it's closed subspace would inherit this property. So if we treat $N$ as subspace and we know it's closed we can say it's compact. But than we know it's divergent series due, to it's property ($d(a_n,a_l) \ge 1/3 $ for any $l \neq n$), so if we pick cover of N by balls centered around it's points with radius lets say... $1/4$, than we can't pick any finite subcover, so $N$ can't be compact, and we have contradiction to our assumption.

Thank you in advance!


For your first step, the idea is correct, but it needs polishing. Speficically, you haven't shown that $r>0$, so you don't know if you can make the open ball around $x$.


Let's say $r=0$. Then there exists some $a_i$ such that $d(x, a_i) < \frac 16$.

Now, try to use triangle inequality to show that $d(x, a_j)<\frac 16$ would lead to a contradiction.

From this, you can conclude that $d(x, a_k)<\frac 16$ for only one value of $k$, which means that $r$ cannot be $0$.

Hint 2:

Think about what the situation would mean. It would mean that $d(x, a_i) < \frac 16$, and also $d(x, a_j)<\frac 16$.

Think about it in words. $x$ is only $\frac 16$ away from two elements, $a_i$ and $a_j$, but the two elements themselves are more than $\frac 13$ away from each other. It's as if city $A$ was $1$ mile from city $B$ and $1$ mile from city $C$, but the distance between cities $B$ and $C$ was $3$ miles. Why is this impossible?

This can also be done without going to proof by contradiction, as you can (without defining $r$ in the first place), prove that

There can be at most one $a_i$ such that $d(x, a_i)<\frac16$.

Using this, you can already conclude that $d(x, N)\neq 0$.

Your second step is entirely correct, sorry for misleading you.

One thing of note:

In $N$, the ball, centered around $a_i$ with a radius of $\frac14$, is exactly the set $\{a_i\}$ (it contains one element).

  • $\begingroup$ With 2'nd step I assume that K is compact by contradiction, and I say that if I treat N as closed subspace of it, than it also should be compact, because proofwiki.org/wiki/Closed_Subspace_of_Compact_Space_is_Compact but it isn't, because if I pick open Balls with radius 1/4 as stated in my answer and cut it to subspace it still will be open ball cover there, but whitout finite subcover. So we have not compact, closed subspace... Or maybe I am missing sth. $\endgroup$ – Kiwi Aug 23 '16 at 13:27
  • $\begingroup$ @Kiwi Ah, yes, then your second answer is completely correct. Sorry about that. $\endgroup$ – 5xum Aug 23 '16 at 13:28
  • $\begingroup$ @ 5xum Sure ^^ No problem! And with 1st one I lack of idea how to show that r>0. I mean, I havn't thought about that problem and now I am a bit stuck. Because if I assume that r=0, than by other definition of closed set in metric space I will have to show that this already is in N, if I am correct. $\endgroup$ – Kiwi Aug 23 '16 at 13:29
  • $\begingroup$ @Kiwi See the edit to my answer! $\endgroup$ – 5xum Aug 23 '16 at 13:36
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    $\begingroup$ @Kiwi I told you you are overcomplicating. The technique I used in Hint 2 is really useful. Take a step back, and try to thing in simple everyday terms. This sometimes lifts the fog of abstract confusion and gives new insight. powodzenia! $\endgroup$ – 5xum Aug 23 '16 at 14:30

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