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I'm working through a book called "Introduction to Topology" and I'm currently working on a chapter regarding metric spaces and continuity. This is how my book defines continuity at a point:

Let $(X,d)$ and $(Y,d')$ be metric spaces, and let $a\in X$. A function $f: X\to Y$ is said to be continuous at the point $a\in X$ if given $\epsilon \gt 0$, there is a $\delta \gt 0$, such that $$d'(f(x),f(a)) \lt \epsilon$$ whenever $x\in X$ and $$d(x,a)\lt \delta$$

My question is this: is it possible that a function may be continuous in the metric spaces $(X,d)$ and $(Y, d')$ but not be continuous if one of the distance functions $d$ or $d'$ is changed to a different distance function?

In other words, does the continuity of a function depend on the distance functions used to "measure" it?

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    $\begingroup$ yes, completely. The topology define the continuity of the functions in the space, and not all distance functions are equivalent. $\endgroup$ – Masacroso Jul 11 '17 at 13:59
  • $\begingroup$ Indeed if it weren't we wouldn't have to define continuity for functions between metric spaces -- we could just define it as a property of sets $X$ and $Y$. If you think about the intuitive meaning of continuity, it's something like "if $x$ is close to $a$ then $f(x)$ is close to $f(a)$". To do this, the concept of "close" needs to make sense. The addition of a metric gives you one way to do that: "close" means "small distance". Putting a topology on the set gives you another way of defining what "close to $a$" means: instead of distance it introduces the idea of "neighbo(u)rhoods" of $a$. $\endgroup$ – CompuChip Jul 11 '17 at 21:11
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Yes. Example: Take the metric where $d(x,x) = 0$ and $d(x,y)=1$ for $x\neq y$ and the Euclid metric. In the first one all functions are continuous and in the second you can find non-continuous functions

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Yes, it does depend on the metric but indirectly.

On a metric space like $(X,d)$ and $(Y,d')$ one defines the topology $\mathcal{T}_d$ or $\mathcal{T}_{d'}$ induced by the metric. This is the smallest topology such that all "open balls" $B(x,r) = \{y \in X: d(x,y) < r\}$ (where $x \in X, r>0$) of the metric are open.

Continuity of $f: (X,d) \to (Y,d')$ depends only on $\mathcal{T}_d$ and $\mathcal{T}_{d'}$; it is a purely topological notion. But we can prove that your criterion is just a reformulation of "$f$ is $(\mathcal{T}_d,\mathcal{T}_{d'})$-continuous at $a$" in terms of the metric that defined the topology.

There are situations where choosing a different metric $d''$ on $X$ will give the same topology so $\mathcal{T}_d = \mathcal{T}_{d''}$, or similarly on $Y$. In that case the metrics are called equivalent. In that case the continuity is not affected by changing the metric.

But if $d$ is the standard metric on $\mathbb{R}$, so $d(x,y) = |x-y|$, and we define another metric $d'(x,y) = |x| + |y|$ (if $x \neq y$ and $d'(x,y) = 0$ otherwise), it turns out that continuity of any function on $(\mathbb{R},d)$ at $0$ is not affected by changing the metric from $d$ to $d'$ (keeping the codomain the same), but at all other points $p \neq 0$ any such function $f$ is always continuous at $p$ in the $d'$ metric. The topologies $d$ and $d'$ are non-equivalent on $\mathbb{R}$: in $d$, $\{1\}$ is not open but in $d'$ it is.

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    $\begingroup$ Pointing out that continuity only depends on the topology is a good idea, but the wording of the question suggests that the asker may not be familiar with the concept, so the answer might benefit from being rewritten at a slightly lower level, including an overview of the definition of topology and its relation to metric spaces. $\endgroup$ – Robin Saunders Jul 11 '17 at 19:09
  • $\begingroup$ Umm, $d'(x, y) = |x| + |y|$ doesn't satisfy $d'(1, 1) = 0$ so it's not a metric. $\endgroup$ – Daniel Schepler Jul 11 '17 at 22:22
  • $\begingroup$ @DanielSchepler yes, we'll have to special case that. Thx for spotting it. $\endgroup$ – Henno Brandsma Jul 11 '17 at 22:23
  • $\begingroup$ @EpsilonNeighborhoodWatch I didn't know one needed reputation for that.. $\endgroup$ – Henno Brandsma Jul 12 '17 at 7:11
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Absolutely it is possible. If you put the discrete metric on $X$ (i.e. the metric defined by $d(x,y) = 1$ when $x \neq y$ and $d(x,x) = 0$) then any function $f:X \to Y$ is continuous because any set in $X$ is open. Thus, for exampl, we can take any function $f:\mathbb R \to \mathbb R$ which is discontinuous with the standard metric on $\mathbb R$ and it will become continuous if we replace the metric on the source space by the discrete metric.

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Yes, continuity depends on the topology.

Intuitively, continuity of $\ f:X \to Y$ means that when you take any point $x \in X$, if some points are "close" to $x$, then the images of those points must be close to $f(x)$.

In a discrete space, all points are far away. In some other topologies, this is not the case.

If $X$ has the discrete topology, then "if some points are close to $x$" never happens, therefore the statement "if some points are close to $x$, then their images are close to $f(x)$" is trivially true (independent of the topology on $Y$).

On the other hand, if $X$ has a non-discrete topology, but $Y$ has the discrete topology, then it means that there are some points that are "close" to $x \in X$. However, those points will never be close to $f(x)$ (unless they are $f(x)$) because $Y$ has the discrete topology.

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Take the spaces $X= \mathbb{R}$ with the discrete metric and the space $Y=\mathbb{R}$ with the euclideian metric $d(x,y)=|x-y|$ the function $f(x)=1$ if $x=0$ and $f(x)=0$ if $x \neq 0$ is continuous using the fact that the inverse image of an open set is open and that in a discrete space every set is open.

But in $X$ if you change the discrete metric with the euclideian metric then $f$ is not continuous.

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