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If $(X,d)$ is a metric space. Is it possible to construct a bijective continuous map $f(X,d)\mapsto \mathbb{R}$. I think it is not possible.Could any one help me to give me hints.

Andy
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    What about $X={0}$? Or any countable metric space, say. – Julien May 02 '13 at 04:09
  • what is $X$, what is $d$? Can they be anything? Can they be $(\mathbb{R},|\cdot|)$? – mathemagician May 02 '13 at 04:09
  • @mathemagician $X,d$ can be anything. – Andy May 02 '13 at 04:13
  • then how about $f(x)=x$? – mathemagician May 02 '13 at 04:14
  • If $X$ is not a subset of $\mathbb{R}$ – Andy May 02 '13 at 04:17
  • @AnindyaGhatak If you ask whether, in general, any metric space can be sent bijectively and continuously onto $\mathbb{R}$, this is obviously wrong. See my comment above. If it is not what you meant to ask, please clarify your question. – Julien May 02 '13 at 04:22
  • @julien I asked for any metric space.could you tell me why it is obviously wrong – Andy May 02 '13 at 04:30
  • Because, for instance, there is no bijection between $\mathbb{R}$ and a countable set. As $\mathbb{R}$ is uncountable. Not even talking about continuity. And there exist countable metric spaces. ${0}$ is an example. $\mathbb{Z}$, or $\mathbb{Q}$, with the usual distance, are less trivial examples. – Julien May 02 '13 at 04:32

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There seem to be two kinds of counterexample to this statement.

The first kind arises from the failure to find some bijective map of sets $X \rightarrow \mathbb{R}$, where $X$ is a set for which some metric $d$ has been defined. This is an issue of cardinality, since here $X$ can really be any set under the sun.

I should point out here that any set possesses at least one metric- $d(x,y) = 1$ when $x \neq y$ and $d(x,x) = 0$. This metrizes the discrete topology on $X$, namely the topology where every point (singleton) is an open set.

The second kind of counterexample arises from the failure to find some continuous bijective map. Suppose, for instance, that I knew ahead of time that $X$ and $\mathbb{R}$ had the same cardinality; I could then ask whether $(X,d)$ could be mapped bijectively and continuously to $\mathbb{R}$. This is not the case, however, as there are no restrictions on the 'wildness' of the metric $d$.

You can create a host of counterexamples, but here's one that's rather damning: let $[0,1]$ and give $X$ the topology $d(x,y) = |x-y|$. Suppose for the sake of contradiction that a continuous, bijective map $f : X \rightarrow \mathbb{R}$ could be found. Then the image of $X$ is a compact set by continuity, and since $f$ is onto, this implies that $\mathbb{R}$ is compact. This is a contradiction. Notice also that we didn't even use injectivity, only surjectivity of the map $f$. I should caution that this isn't a great counterexample, however, because to be fair there are far more fundamental reasons why this should not be the case without stringent conditions on $(X,d)$.

A Blumenthal
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