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Let $f$ be the identity function from $(X,d_1) \to (X,d_2)$. If $d_1$ and $d_2$ are equivalent metrics we can deduce that the identity function is continuous, right?

Since for every open set $G$, its pre-image under $f$ would also be $G$, and since the metrics are equivalent an open set in one metric is open in the other?

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    $\begingroup$ Yes, this is correct. $\endgroup$ – Brian M. Scott Jul 3 '13 at 5:43
  • $\begingroup$ Yes. In fact, the identity function is a homeomorphism. $\endgroup$ – nigel Jul 3 '13 at 5:54
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Yes. In fact, the following criterion is useful:

The identity function $f:(X,d_1)\to (X,d_2)$ is continuous if and only if for all $x\in X$ and for all $\epsilon>0$, there exists $\delta>0$ such that $x\in B_{d_1}(x,\delta)\subseteq B_{d_2}(x,\epsilon)$. ($B_{d_i}(x,r)$ denotes the ball of radius $r$ centered at $x$ with respect to the metric $d_i$ on $X$.)

In particular, the condition that the identity function be continuous is weaker than the condition that $(X,d_1)$ and $(X,d_2)$ be equivalent metric spaces. The latter condition is equivalent to the assertion that the identity function and the inverse of the identity function are continuous.

Exercise 1: Prove the criterion stated above using the definition of continuity via open sets.

Exercise 2: Let $d_1$ be the metric on $\mathbb{R}$ defined by the rule $d_1(x,y)=1$ if $x\neq y$ and $d_1(x,x)=0$ for all $x,y\in\mathbb{R}$. Let $d_2$ be the standard metric on $\mathbb{R}$. Prove that the identity function $f:(\mathbb{R},d_1)\to (\mathbb{R},d_2)$ is continuous. However, prove that the identity function $g:(\mathbb{R},d_2)\to (\mathbb{R},d_1)$ is not continuous!

Exercise 3: Let $d_1$ be the standard metric on $\mathbb{R}$ and let $d_2$ be the metric defined by the rule $d_2(x,y)=\text{min} \{1,d_1(x,y)\}=\text{min} \{1,\left|x-y\right|\}$ for all $x,y\in \mathbb{R}$. Prove that $d_1$ and $d_2$ are equivalent metrics on $\mathbb{R}$ and verify independently that both the identity function $f:(\mathbb{R},d_1)\to (\mathbb{R},d_2)$ and its inverse $g:(\mathbb{R},d_2)\to (\mathbb{R},d_1)$ are continuous.

I hope this helps!

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Yes.

To say that the metrics $d_1$ and $d_2$ are equivalent amounts to saying that, if ${\mathcal T}_1$ and ${\mathcal T}_2$ are the topologies generated by them, then ${\mathcal T}_1 = {\mathcal T}_2$.

Let $\mathcal{T}_1$ and $\mathcal{T}_2$ be two topologies on some space $X$. (For example $\mathcal{T}_1$ and $\mathcal{T}_2$ could be the topologies generated by the metrics $d_1$ and $d_2$, respectively, but the points made below apply more generally.)

Also, let $I$ be the map $(X, \mathcal{T}_1)\to (X, \mathcal{T}_2)$ defined by $x \mapsto x$. 1

Now, the following four statements are absolutely equivalent; they differ only in clarity or directness. Take your pick:

  1. $\mathcal{T}_2 \subseteq \mathcal{T}_1$;
  2. for all $U \in \mathcal{T}_2$, the set $I^{-1}(U) = U$ belongs to $\mathcal{T}_1$;
  3. for any open set $U$ in $(X, \mathcal{T}_2)$, the preimage $I^{-1}(U)$ is open in $(X, \mathcal{T}_1)$;
  4. the map $I$ (as defined above) is continuous.

Now, clearly, the map $I$ is bijective, and therefore its inverse $I^{-1}:(X, \mathcal{T}_2)\to (X, \mathcal{T}_1)$ is a well-defined map. This means that the duals of the four statements above are also equivalent to each other; these four equivalent statements are:

  1. $\mathcal{T}_1 \subseteq \mathcal{T}_2$;
  2. for all $U \in \mathcal{T}_1$, the set $I(U) = U$ belongs to $\mathcal{T}_2$;
  3. for any open set $U$ in $(X, \mathcal{T}_1)$, the the image of $U$ under $I$ is open in $(X, \mathcal{T}_2)$;
  4. the inverse $I^{-1}$ of the bijection $I$ (as defined above) is continuous.

Next, we have the following two equivalent statement, the first obtained by combining the first statements of the two sets above, and the second by combining the last ones:

  1. $\mathcal{T}_1 = \mathcal{T}_2$;
  2. the bijection $I$ (given by $x \mapsto x$) is homeomorphic.

And, as already pointed out before, in the special case where ${\mathcal T}_1$ and ${\mathcal T}_2$ are the topologies generated by the metrics $d_1$ and $d_2$, respectively, the last two statements are also equivalent to:

  3. the metrics $d_1$ and $d_2$ are equivalent.

So, as others have pointed out, the equivalence of the two metrics renders the map $x\mapsto x$ not only continuous, but in fact homeomorphic.


1 I'm refraining from calling $I$ the "identity" out of pedantry... Strictly speaking, the domain and codomain of the identity are the same, while here we're treating the domain $(X, \mathcal{T}_1)$ and codomain $(X, \mathcal{T}_2)$ of $I$ as distinct topological spaces (even if they both have the same "underlying set" $X$). The "function of sets underlying" the map $I$ is certainly the identity $X \to X$. I don't like all this fuss more than anyone else, but when one starts talking about "equivalent metrics", etc., it's difficult to escape such pedantry altogether.

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  • $\begingroup$ The question was more specifically about metrics. The topological story was also discussed here. $\endgroup$ – wildildildlife Jul 8 '13 at 11:34

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