Showing a metric space is not complete by showing the set is neither open nor closed

complete-spacesmetric-spaces

I have to determine whether a metric space $(X,d)$ where $X=\{(x,y,z) \in \mathbb R^3 : x^2+y^2+z^2<1, x\geq 0\}$ and $d$ is the induced metric from $\mathbb R^3$ (which I believe is $d_2 = (\sum_{j=1}^k |x_j-y_j|^2)^{\tfrac{1}{2}}$) is complete or not.

The approach I am trying to take is to show that $X \subset \mathbb R^3$ is neither open nor closed in $\mathbb R^3$, which is complete with respect to the $d_2$ metric and as such, it fails the completeness characterization for subsets theorem which states that if a metric space $(X,d)$ is complete, then a subset $A \subset X$ is complete with respect to the induced metric if and only if $A$ is closed.

I am still not sure that even if I do manage to show that it is neither open nor closed, I can actually apply the theorem since I have no clue whether it applies to sets which are neither (I haven't been able to find an answer for that either, but I digress). The lemma we were shown to prove such a set is closed in another is to show that for every sequence $x_n \in A$ converging in $X$, one also has that the $lim_{n \to \infty} (x_n) \in A$. My "proof" does not make use of this lemma. It is as follows:

Consider some point $x_1=(0,\tfrac{1}{2},0)$. Then, $\forall \epsilon>0, B_{\epsilon}(x_1)$ (ball with radius $\epsilon$ centered on the point $x_1$) contains the point $(0-\tfrac{1}{2}\epsilon,\tfrac{1} {2},0) \notin X$ showing that $B_{\epsilon}(x_1) \not\subset X \implies X$ is not open. A similar argument is used to show that the complement $\mathbb R^3 \setminus X$ is not open $\implies$ $X$ is not closed. As such, it fails the aforementioned theorem for the completeness characterization for subsets (I think?) and as a result, $(X,d)$ is not complete.

I also know that I did not make use of the metric space definition of open balls. Intuitively, I am aware how that definition applies, but I am not able to formalize it in this context. Definition is as follows; For any metric space $(X,d_X)$, the open ball with centre $x \in X$ and radius $r$ is the subset $B_r(x) = \{y \in X : d_X(y,x) < r\}$.

I am not entirely confident with my approach and any hints are much appreciated. Thank you in advance!

Best Answer

The theorem you are trying to apply is:

if a metric space $(X,d)$ is complete, then a subset $A \subset X$ is complete with respect to the induced metric if and only if $A$ is closed.

Now comes a somewhat confusing part: In your case the "$X$" of the theorem is $\Bbb R^3$, while the "$A$" of the theorem is your set "$X$". Rather than maintain this mix up of the meaning of "$X$", let's just relabel that variable in the theorem:

if a metric space $(S,d)$ is complete, then a subset $A \subset S$ is complete with respect to the induced metric if and only if $A$ is closed.

The implications here are clear: $A$ is complete if and only if $A$ is closed.

  • If $A$ is complete, then $A$ is closed. ("only if" clause)
  • If $A$ is closed, then $A$ is complete. ("if" clause)
  • If $A$ is not complete, then $A$ is not closed. (contrapositive of "if" clause)
  • IF $A$ is not closed, then $A$ is not complete. (contrapositive of "only if" clause)

Now in your application, $S = \Bbb R^3$, and $A = X$.

I am still not sure that even if I do manage to show that it is neither open nor closed, I can actually apply the theorem since I have no clue whether it applies to sets which are neither (I haven't been able to find an answer for that either, but I digress).

The answer is right there in the theorem. If $X$ is not closed, then it is not complete.

"Open" has nothing to do with it. It doesn't matter whether $X$ is open or not. All you need to do is show that $X$ is not closed in $\Bbb R^3$, and the theorem implies that $X$ is not complete.

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