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Which of the following is/are true?

  1. (0,1) with usual topology admits a metric which is complete.
    1. (0,1) with usual topology admits a metric which is not complete.
    2. [0,1] with usual topology admits a metric which is not complete.
    3. [0,1] with usual topology admits a metric which is complete.

Since every subspace of a metric space is complete iff it is closed. So R taken as topological space with usual topology and (0,1) &[0,1] as subspace of R .then (0,1) is open and [0,1] is closed. So options should be true 2&4. But options are true 1,2&4.(According to CSIR)

Also I'm seeing 1&2 are negation of each other. Where is my misunderstanding? Please convince me!! I will appreciate any efforts by you. Thanks in advance.

Mauro ALLEGRANZA
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user697748
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    If you think #1 and #2 are negations of each other than you haven't understood them. The two metrics involved would not both be the same metric; rather they would only both induce the same topology. #1 and #2 are both true. Furthermore, as a subspace of itself $(0,1)$ is both open and closed. – Michael Hardy Oct 23 '19 at 19:24

3 Answers3

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$3$ is false due to $[0,1]$ being compact in the usual topology (see Every compact metric space is complete), and $2$ and $4$ are true for the reason you mention (with the metric inherited from $\mathbb R$).

Now to $1$: We could imbue $(0,1)$ with a new metric, not inherited from $\mathbb R$, in which it will be complete, yet which will produce the same topology as usual. Such a metric space will be a subspace of $\mathbb R$ as a topological space but not as a metric space. (Thus the theorem you cited doesn't apply.)

One such example would be $d(x,y)=|f(x)-f(y)|$ where $f(x)=\tan\left(\pi\left(x-\frac{1}{2}\right)\right)$. That would make $(0,1)$ isometric with $\mathbb R$ in the usual sense, and therefore it will be complete.

(Note: $f$ could have been chosen as any homeomorphism of $(0,1)$ to $\mathbb R$, the 'tangent' function is merely a convenience.)

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  1. True. Take for instance an homeomorphism $T:(0,1)\to \mathbb{R}$ and pull back the metric on $\mathbb{R}$ by setting $d(x,y) = |T(x)-T(y)|$.
  2. True. Usual metric.
  3. False. As soon as $[0,1]$ is metrized, compactness is equivalent to completeness and total boundedness.
  4. True. See point 3.

I do not quite get your flow of thoughts, so I cannot answer where is your misunderstanding. Anyways, completeness has nothing to do with the topology. To clarify a bit why both point 1 and 2 are true, think of the following: if you are just given a space (forget that $(0,1)\subset \mathbb{R}$), then there is nothing that tells you that you are reaching some boundary, because this does not exists in itself! So sequences that converge to zero can be seen also as sequences that diverge.

Tommaso Seneci
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Being complete isn't a topological property, but a property of the metric space. Now, 1 only asks for some metric $d$ on $(0,1)$ which induces the usual topology and makes the metric space complete. And this can be seen easily: Take some homeomorphism $(0,1) \to \mathbb R$ and pull back the metric from $\mathbb R$ to $(0,1)$.