which of the following is/are true ?

  1. $(0,1)$ with the usual topology admits a metric which is complete .

  2. $(0,1)$ with the usual topology admits a metric which is not complete.

  3. $[0,1]$ with the usual topology admits a metric which is not complete.

  4. $[0,1]$ with the usual topology admits a metric which is complete.

This question has come at my competetive exam. I think this is a wrong question, because completeness a metric space property not a topological space property.In the offical answer key, answer has given (1) and (4), I want to send my representation. So please check my representation. Thank you

Let $X = (0,1)$ and $d$ is a Euclidean metric on $X$ which induces the usual topology on $X$ and a sequence $\{\frac{1}{n} \}$ is a cauchy sequence in the Euclidean metric , but not converges in $X$. So $X$ is not complete withbthe usual topology admit a Euclidean metric.

On the other hand

The map $$ f:(0,1)\to\mathbb{R}:x\mapsto\tan\pi\left(x-\frac{1}{2}\right) $$ is a bijection which allows you to define the metric $$ d(x,y)=|f(x)-f(y)| $$ which makes $((0,1),d)$ complete. Since $f$ maps intervals to intervals then both topologies are equivalent.

So Completeness is not a topological property. So this is irrelivent.

I would be thankful, if some one check my representation

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  • 1
    $\begingroup$ Um, the question doesn't say "complete topology"; it says "complete metric". So there's nothing wrong with the questions. $\endgroup$ – fleablood Jul 18 '16 at 4:02
  • 1
    $\begingroup$ $ 2$ and $4$ are clearly true. $\endgroup$ – Jorge Fernández Hidalgo Jul 18 '16 at 4:03
  • $\begingroup$ The map f is irrelevant as that admits a different metric. The questions are specifically asking about the euclidean metric. $\endgroup$ – fleablood Jul 18 '16 at 4:05
  • $\begingroup$ @ fleablood :There are two metric induced the same topology and one is complete other is incomplete. So 1 and 2 are true. $\endgroup$ – user120386 Jul 18 '16 at 4:07
  • $\begingroup$ In the offical answer key , answer is given 1 and 4 $\endgroup$ – user120386 Jul 18 '16 at 4:09

I think it is a legit question to ask if a topological space admits a metric.

If I give you a topological space and ask you if there is a metric (with certain properties) which induces this topology there is nothing wrong about it, right? (If I got the word admit right...)

It is right that completeness may does not make sense on any space, but if I give you a space I can ask you if it makes sense. There is a broader class of spaces in which it make sense. If you are interested you can read about uniform spaces.

Now why are 1 and 4 correct? The space in 4 is a closed subspace of a complete space ($\Bbb R$) which is complete. And for 1 you need that $(0,1)$ is homeomorphic to $\Bbb R$ which is complete.

Clearly to show that 2 is correct you can show that $(0,1)$ with the usual topology is not complete, just choose a Cauchy sequence which converges to $0$.

Most interesting is 3. You have to know that $[0,1]$ with any metric inducing the standard topology is complete. Here you can use the fact that any compact metric space ist complete, since compactness is clearly a topological property.

So either I got the question wrong or 2 is also correct. In the second case you can have a look at Completely metrizable space which are topological spaces whose topology is/(can be) induced by a complete metric.

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