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It is proven that any metric $d$ is continuous. Consider the metric space $(\mathbb{R}, d)$ where:

$$d:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$$

$$d(x, y) = \begin{cases} 0 & x=y \\ 1 & x\neq y \end{cases} $$

Let $x_n \rightarrow x=0$ and $y_n \rightarrow y=0$. If you take $d(x_n, y_n)\rightarrow 1 \neq 0=d(x, y)$. This shows that this metric is discontinuous. What is wrong with my reasoning?

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  • $\begingroup$ $x_n=0=y_n$ for all $n$ large. $\endgroup$
    – Jo'
    Oct 7, 2018 at 15:25
  • $\begingroup$ You write $\mathbb{R}$ but it has two different topologies for this statement. See my answer. $\endgroup$ Oct 7, 2018 at 21:30

3 Answers 3

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The fact that $d(x_{n},x)\to 0$ implies that eventually $d(x_{n},x)<1/2$. Thus, eventually the $x_{n}$'s must all be $0$. The same goes for the $y_{n}$'s. Therefore, $d(x_{n},y_{n})$ is eventually $0$ for such sequences.

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You misrepresent the statement.

Correct is: let $(X,d)$ be a metric space. This induces a topology $\mathcal{T}_d$ on $X$ induced by the metric (the minimal topology where all balls $B(x,r)$ are open). Then the map $d: (X,\mathcal{T}_d) \times (X,\mathcal{T}_d) \to \mathbb{R}$ is continuous where the left hand side has the product topology of the metric topology with itself and the reals have the usual Euclidean topology.

If we use the discrete metric, which induces the discrete topology, then this product topology is also discrete and $d$ is indeed continuous (as is any map on a discrete space, so it's not very informative).

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A sequence $x_n$ has limit $0$ in the discrete metric if and only if eventually $x_n = 0$. Does this help?

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  • $\begingroup$ Why is that so? Isn't $x_n$ a normal sequence such as $\frac{1}{n}$? In which case $x_n\neq 0 \forall n$? $\endgroup$ Oct 7, 2018 at 15:26
  • $\begingroup$ Ok, understood what you mean with the help of ervx's answer :) thanks $\endgroup$ Oct 7, 2018 at 15:29

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