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Consider $f, g \in C([0,1])$ There are two different metrics: $d_1(f,g)=\max_{x \in [0,1]}|f(x)-g(x)|$ and $d_2(f,g)=\inf[\epsilon \geq 0; X\subset Y_\epsilon$ AND $Y\subset X_\epsilon ]$ where $X=\operatorname{graph}(f)$ and $Y=\operatorname{graph}(g)$ Note here that $X_\epsilon$ is the set of all points within $\epsilon$ of $X$.

I have encounted a claim that $(C([0,1], d_1)$ is a complete metric space, but that $C(0,1),d_2)$ is not complete. I am failing to see how to see one can be complete but not the other. It seems to me that these metrics are equivalent, so I think feel that either both metric spaces should be complete or both not. Any tips or hints would be much appreciated.

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  • $\begingroup$ When you say that $X_\epsilon$ is the set of all points within $\epsilon$ of $X$, do you mean as subsets of $\mathbb{R}^2$? If so, consider $f(x)=x$ and $g(x)=x^2$. You'll find that $d_1(f,g)=1/4$, while $d_2(f,g)=1/\sqrt{32}$, so these metrics are not the same. $\endgroup$ – 211792 Dec 15 '17 at 20:13
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    $\begingroup$ @AustinChristian Metrics can be equivalent even if they are not the same.Two metrics are (topologically) equivalent if they generate the same topology, i.e. the same set of open sets. Thus the notion of convergence is the same. Please see en.wikipedia.org/wiki/Equivalence_of_metrics $\endgroup$ – user491874 Dec 15 '17 at 20:17
  • $\begingroup$ @user8734617 You're right. For some reason I thought that the OP thought the two metrics were the same (and not just equivalent). $\endgroup$ – 211792 Dec 15 '17 at 20:18
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    $\begingroup$ For metric spaces a more useful notion of equivalence is obtained when one requires that both identity functions $i:(M, d_1) \rightarrow (M, d_2)$ and $i:(M, d_2) \rightarrow (M, d_1)$ are uniformly continuous instead of just continuous. $\endgroup$ – lzralbu Dec 15 '17 at 20:45
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For $n\in\mathbb N$, $n\gt 2$, take:

$$f_n(x)=\begin{cases}0 & 0\le x\le\frac{1}{2}-\frac{1}{n}\\ 1 & \frac{1}{2}+\frac{1}{n}\le x \le 1 \\ \text{linear} & \frac{1}{2}-\frac{1}{n}\le x \le \frac{1}{2}+\frac{1}{n}\end{cases}$$

In other words, $f_n(x)$ has a graph which is a polygonal line $ABCD$ where $A=(0, 0)$, $B=(\frac{1}{2}-\frac{1}{n}, 0)$, $C=(\frac{1}{2}+\frac{1}{n}, 1)$ and $D=(1,1)$

Now this is a Cauchy sequence wrt. $d_2$ that does not converge. (Proof: exercise.)

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