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Is it possible to have a metric space M such that there is no continuous map from M to any other metric space?

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This question together with your username iamgroot reminded me of rigid spaces. A topological space $X$ is called rigid if the only homeomorphism $X\to X$ is the identity map. IIRC also some rigid continua have been constructed. This topic was studied (among others) by Johannes de Groot. –  Martin Sleziak Aug 13 at 5:44

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No, assuming that the spaces are non-empty. In fact take any two non-empty topological spaces $X$, $Y$. Pick your favorite element of $Y$ and call it $y_0$. Put $$ f : X \to Y \\ x \mapsto y_0 $$ Now we have for any open set $A \subset Y$ that $f^{-1}(A) = X$ or $f^{-1}(A) = \emptyset$ depending on if $y_0 \in A$. Of course since $X$ is a topological space we have that both $X, \emptyset$ are open sets.

If you want to restrict this to just metric spaces and use the definition of continuity that may be more familiar when working with metric spaces, consider two metric spaces $X,Y$, put your favorite element of $Y$ to be $y_0$ and put $$ f : X \to Y \\ x \mapsto y_0 $$ Let $\epsilon > 0$. Take $\delta$ to be your favorite positive real number. Now we have $$ d_X(x,y) < \delta \implies d_Y(f(x),f(y)) = d_Y(y_0, y_0) = 0 < \epsilon $$ so indeed $f$ is continuous.


EDIT: I wanted to take some time to talk about the cases where $X,Y$ are possibly empty (mostly for pedagogical reasons).

First let's assume $X$ is non-empty but $Y$ is empty. Then there doesn't exist a map $f : X \to Y$, so in this case there actually is no continuous function from $X$ to $Y$.

Next if $X$ is empty but $Y$ is non-empty then there is a unique empty function $f_Y : X \to Y$ which is going to be continuous (since $f^{-1}(A) = \emptyset$ is open for any open $A$).

Lastly if both $X$ and $Y$ are empty then we again have the unique empty function $f_Y : X \to Y$ which is as well continuous since $f^{-1}(Y) = X$.


EDIT$^2$ Just realized the specific question was if there exists a metric space $M$ that doesn't have a continuous map to any other metric space. Based on what I said above it should be easily be seen that this cannot be true since we can just take our favorite non-empty metric space and proceed as above.

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you need to assume $Y$ is not empty. –  Ittay Weiss Aug 13 at 4:16
    
In other words: constant maps are always continuous in any topological setting (metrizability not required). You can do a little better: the same argument shows that maps that are constant on each component of the domain are always continuous (but there may only be one component, so this may not add much). –  MPW Aug 13 at 4:17
    
@IttayWeiss yes good call - originally had it in there but forgot to re-add it :P –  DanZimm Aug 13 at 4:18
    
@MPW what do you mean by "component of the domain"? –  DanZimm Aug 13 at 4:19
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@IttayWeiss: I don't think you have to make that assumption. Every map from any topological space $X$ into $\varnothing$ is continuous because there is no such map if $X\neq\varnothing$, and $f^{-1}(\varnothing)=\varnothing$ is open otherwise. –  MPW Aug 13 at 4:21

Take any non-empty subset $A$ of a metric space $(X, d)$. Let $d(x, A): X\to\mathbb{R}$ be defined by $d(x, A)=\operatorname{inf}\{d(x, A): x\in X\}$. Then you can check that $d(x, A)$ is a continuous function from $X$ to $\mathbb{R}$. In fact, it is uniformly continuous.

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