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Metric Space: (M,d)

Set:

$M = \{ (x,y)\in \mathbb{R}^2:y>0 $ or $ x=0=y \}$

Metric:

$d((x,y),(a,b)) = $min$\{ $max$ \{ |x-a|,|y-b| \},y+b \}$

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completeness:

$\lim_{n \to \infty}(1,\frac{1}{n}) = (1,0)$ which does not exist in the set, hence the metric space is not complete

compactness:

The sequence $a_n = (n,n)$ contains no subsequence which converges in $M$. Hence the space is not compact.

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Are my answers correct? If so, what was the purpose of the distance function? If not, where am I going wrong? Thanks for any answers.

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  • $\begingroup$ $a_n = n$ is not a sequence in this set. $\endgroup$ – Patrick Stevens Dec 31 '15 at 18:28
  • $\begingroup$ Is $a_n = (n,n)$ correct? $\endgroup$ – Gregory Peck Dec 31 '15 at 18:29
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Be careful! In the usual metric, we do indeed have that $(1,\frac1n)$ converges to $(1,0),$ but we aren't using the usual metric. In fact, the sequence does converge in this space, to the point $(0,0)$! In fact, given any sequence of points of $M$ that we would expect to converge somewhere on the $x$-axis, it will converge to $(0,0)$ in $\langle M,d\rangle.$ This is even true of sequences that approach the $x$-axis that we wouldn't expect to converge, like $$v_n:=\left(n,2^{-n}\right).$$ Aside from that, convergence behaves the way we'd expect it to. Hence, the space is complete!

To prove it, take any Cauchy sequence. If the sequence is bounded away from the $x$-axis--that is, if the second components don't converge to $0$--then you can show that the sequence converges where we'd expect it to converge. Otherwise, you can show that the sequence converges to $(0,0),$ despite our expectations.

Your approach to prove it isn't compact is a good one, but you still have to show that no subsequence can converge. Probably the simplest approach is to show that no subsequence is Cauchy.


Added: Let me expand on my answer, since this metric has some strangely-shaped open "balls." Let's denote and define the open "ball" with "radius" $r$ and "center" $(h,k)$ by $$B_r(h,k):=\Bigl\{(x,y)\in M:d\bigl((x,y),(h,k)\bigr)<r\Bigr\}.$$


First, let's consider the open "balls" that are "centered" at $(0,0).$ Note that $$d\bigl((x,y),(0,0)\bigr)=\min\Bigl\{\max\bigl\{|x-0|,|y-0|\bigr\},y+0\Bigr\}=\min\Bigl\{\max\bigl\{|x|,y\bigr\},y\Bigr\}.$$ Note that if $|x|<y,$ then $$d\bigl((x,y),(0,0)\bigr)=\min\{y,y\}=y,$$ and if $y\le|x|,$ then $$d\bigl((x,y),(0,0)\bigr)=\min\bigl\{|x|,y\bigr\}=y.$$ Hence, for all $r>0$ we have $$B_r(0,0)=\bigl\{(x,y)\in M:y<r\bigr\}.$$ Consequently, given any sequence of points $(x_n,y_n)\in M$ with $\lim_{n\to\infty}y_n=0,$ we have that $\lim_{n\to\infty}d\bigl((x_n,y_n),(0,0)\bigr)=0.$ That is, any such sequence converges to $(0,0)$ in $\langle M,d\rangle.$


Now let's consider what happens when $(h,k)\ne(0,0).$ Note in particular that this means $k>0,$ given the definition of $M.$

First, let's consider the (relatively) simple case that $0<r\le k.$ Note that $d\bigl((x,y),(h,k)\bigr)<r$ if and only if $\max\bigl\{|x-h|,|y-k|\bigr\}<r$ or $y+k<r.$ (Do you see why?) But $y\ge0$ and $r\le k,$ so $y+k<r$ isn't possible! Thus, $d\bigl((x,y),(h,k)\bigr)<r$ if and only if $\max\bigl\{|x-h|,|y-k|\bigr\}<r,$ which occurs if and only if $|x-h|<r$ and $|y-k|<r.$ (Do you see why?) Thus, $B_r(h,k)$ is the interior of the square with vertices $(h\pm r,k+r)$ and $(h\pm r,k-r),$ in the case that $0<r\le k.$

But what if $r>k$? Well, one can show that $$B_r(h,k)=\bigl\{(0,0)\bigr\}\cup\bigl\{(x,y)\in M:0<y<r-k\bigr\}\cup\bigl\{(x,y)\in M:h-r<x<h+r\text{ and }r-k\le y<k+r\bigr\}.$$ What a mess! For example, here is the rough shape of the open "ball" of "radius" $5$ with "center" $(1,2).$ Fortunately, as far as convergence is concerned, this doesn't really matter! We can restrict ourselves to the case that $0<r\le k$ to show that, given a sequence of points $(x_n,y_n)\in M$ and a point $(h,k)\in M$ with $k\ne 0,$ we have that the sequence converges to $(h,k)$ in $\langle M,d\rangle$ if and only if it converges to $(h,k)$ in the usual metric.

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    $\begingroup$ I believe the following is true: Let $X$ be a metrizable space. If every metric on $X$ is complete then $X$ is compact. So for a noncompact metric space clearly you must pay attention to the specific metric! $\endgroup$ – Forever Mozart Dec 31 '15 at 20:04
  • $\begingroup$ "In fact, given any sequence of points of $M$ that we would expect to converge somewhere on the $x$-axis, it will converge to $(0,0)$ in $⟨M,d⟩$." Could you demonstrate this in more detail. I can't seem to figure it out for myself, I'm very new to metric spaces. $\endgroup$ – Gregory Peck Jan 4 '16 at 17:51
  • $\begingroup$ The kicker is to use the fact that for any $(x,y)\in M,$ we have $$0\leq d\bigl((x,y),(0,0)\bigr)\leq y+0=y.$$ Now consider any sequence of points $(x_n,y_n)$ in $M$ such that $\lim_{n\to\infty}y_n=0.$ $\endgroup$ – Cameron Buie Jan 4 '16 at 18:26
  • $\begingroup$ Ok, that makes sense, I think I've successfully proved that. Now I'm stuck at this: "if the second components don't converge to 0 then you can show that the sequence converges where we'd expect it to converge." This is a very tricky one, and given this metric it doesn't seem intuitive to me. Also, since all subsets of $R^n$ which are bounded are totally bounded, can't we simply show that $M$ is not bounded, and hence is not totally bounded and not compact? $\endgroup$ – Gregory Peck Jan 4 '16 at 20:20
  • $\begingroup$ @Gregory: I will expand my answer to address the convergence. As for boundedness, since compact implies totally bounded, which in turn imples bounded, then it does suffice to show that $M$ isn't an open $d$-ball of finite radius. Still, I think your first approach is better, since you'll need a pretty good handle on the metric for a boundedness proof. $\endgroup$ – Cameron Buie Jan 4 '16 at 23:18

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