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I'm studying Analysis on Manifolds by Munkres, and at page 199, it is given that

Let $S$ be a subset of $\mathbb{R}^k$; let $f: S \to \mathbb{R}^n$. We say that $f$ is of class $C^r$, on $S$ if $f$ may be extended to a function $g: U \to \mathbb{R}^n$ that is of class $C^r$ on an open set $U$ of $\mathbb{R}^k$ containing $S$.

It is clear from this definition that, even if we were working on a subspace $M$ of $\mathbb{R}^n$ (or on the set $M$ with different topology other subspace topology), we still consider the opens sets of $M$ as a subset of $\mathbb{R}^n$, and show the differentiability according to that.

However, for example, if we were to show the continuity of a function $f : \mathbb{R}^k \to M$, we would consider open sets of $M$, as open sets of $M$, i.e not $\mathbb{R}^n$. In this sense, the continuity of a function is depends on the topology of the domain & codomain of that function, whereas the differentiability does not, as far as I have seen.

So my question is that, is there any definition of differentiability that is dependent on the underlying topologies of domain & codomain ?

Clarification

For example, normally, for $f: A \to \mathbb{R}^m$, the concept differentiability is defined for $x \in Int(A)$, but the very definition of interior needs the definition of what we mean by an open set, which is dependent on the underlying topology, so say (trivially) $A = \{1,2\}$, then with the discrete topology both $1,2 \in Int(A)$, but can we define differentiability in this space ?

Or let say, $A = (0,1]$ as a subspace of $[0,1]$ with the standard topology (subspace topology inherited from $\mathbb{R}$).Now if we consider our bigger space as $[0,1]$, then we should be able to define differentiability at $x = 1$ because $(0,1]$ is open in $[0,1]$, hence $1\in Int[0,1]$.

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    $\begingroup$ Look up exotic spheres. There exist, for example, differentiable manifolds, both homeomorphic to $S^7$, but not diffeomorphic to each other. Not really you question, but related. Anyway, when you get to the definition of a differentiable manifold you are close. The idea is that you cover a topological space with open subsets, homemorphic to open subsets of $\Bbb{R}^n$, and use those to locally define what differentiable means. You need a compatibility condition for the intersections. $\endgroup$ – Jyrki Lahtonen Aug 13 '18 at 8:09
  • $\begingroup$ Answer to your clarification is still no as I mentioned in my answer. Differentiability requires a normed space and $\mathbb R^m$ equipped with the discrete topology is not a normed space. $\endgroup$ – mathcounterexamples.net Aug 13 '18 at 11:42
  • $\begingroup$ @mathcounterexamples.net Then, when we have a normed space (any space really), when we say open set, should we consider open sets that are coming from the metric topology on that space ? $\endgroup$ – onurcanbektas Aug 13 '18 at 12:08
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The notion of differentiability is not involving only a topology but a normed space. And in $\mathbb R^m$ all the norms are equivalent.

Therefore the topology used in the definition of differentiability is the one induced by the norm. This topology is unique in $\mathbb R^n$. This maybe different in infinite dimensional Banach spaces.

Finally, there is only one relative topology on a subset $S \subseteq \mathbb R^m$ induced by the « normed topology » of $\mathbb R^m$.

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  • $\begingroup$ what about in a subspace of $\mathbb{R}^n $ ? $\endgroup$ – onurcanbektas Aug 13 '18 at 8:08
  • $\begingroup$ That is not different for subspaces. The relative topologies are induced by topologies defined by equivalent norms. Therefore, there is only one relative topology. $\endgroup$ – mathcounterexamples.net Aug 13 '18 at 8:13
  • $\begingroup$ yes, but I haven't seen a definition of differentiability that considers $[0,1]$ as the bigger space - let $U$ be open in $[0,1]$ - so what is the reason for that ? what is special about $\mathbb{R}^n$ ? why are we never considering differentiability in a subspace of $\mathbb{R}^n $ ? $\endgroup$ – onurcanbektas Aug 13 '18 at 8:17
  • $\begingroup$ The question concerns differentiable manifolds. In a differentiable manifold, there might co-exist more differential structures on the same topology, in contrast to what happens in normed spaces. $\endgroup$ – Giuseppe Negro Aug 13 '18 at 8:41
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    $\begingroup$ @GiuseppeNegro That doesn’t make any difference. Relative topology is defined for subsets, not only subspaces. $\endgroup$ – mathcounterexamples.net Aug 13 '18 at 8:49
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Take a look also at No topologies characterize differentiability as continuity by Geroch, Kronheimer and McCarty (1976).

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    $\begingroup$ a) you should always add a summary of the linked content to a post on StackExchange b) does that paper really address the question? $\endgroup$ – leftaroundabout Aug 13 '18 at 10:22
  • $\begingroup$ @JoseBrox, see my edit also. $\endgroup$ – onurcanbektas Aug 13 '18 at 10:33
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Firstly there's a small error in your original question when you say

to show the continuity of a function $f : \mathbb{R}^k \to M$, we would consider open sets of $M$, as open sets of $M$, i.e not $\mathbb{R}^n$.

Technically we are considering the open sets of $M$ but since $M \subseteq \mathbb{R}^n$ and $M$ has the subspace topology inherited from $\mathbb{R}^n$, so the topology of $\mathbb{R}^n$ plays a vital role in determining continuity of $f$.

If I understand your question correctly, there are two parts to it. The first part is can we only rely on the topology of a set to define differentiability? In other words can we define differentiability for maps between arbitrary topological spaces $X$ and $Y$? And the answer to that is no. (see mathcounterexamples.net's answer)

The second part I think has a simple answer. Some authors (expecially for texts on multivariable calculus) only define differentiability for functions $f : U \to \mathbb{R}^n$ where $U$ is an open susbet of $\mathbb{R}^k$. Now I think from this your misunderstanding is that you're not sure whether to take the interior with respect to the subspace topology or the larger topology.

For example you can extend the definition above to an arbitrary set $S \subseteq \mathbb{R}^k$ and then define differentiability of a function $f : S \to \mathbb{R}^n$ on $$\operatorname{Int}_{\mathbb{R}^k}(S)$$ which by definition of the interior is the largest open set in $\mathbb{R}^k$ contained in $S$ (as opposed to $\operatorname{Int}_{S}(S) = S$ which is the largest open set in $S$ with the subspace topology contained in $S$ which is just $S$). If I recall correctly this is what Munkres did in his book.

As a quick application to the example you gave above. By the above definition, the function $f : (0, 1] \to \mathbb{R}^m$ would only be differentiable on $$\operatorname{Int}_{\mathbb{R}}\left((0, 1]\right) = (0, 1)$$


Why do we only define differentiability for open sets in $\mathbb{R}^n$?

To answer this question you have to go back to the definition of a limit of a function (because differentiability is defined through limits of functions)

Definition: [Limit of a function] Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. Suppose that $E \subseteq X$. Let $f : E \to Y$ and suppose that $p$ is a limit point of $E$. We say that $$\lim_{x \to p} f(x) = q$$ if there exists a point $q \in Y$ with the property that for all $\epsilon > 0$ there exists a $\delta > 0$ such that $$0 < d_X(x, p) < \delta \implies d_Y((f(x), q) < \epsilon$$ for any $x \in B_{(X, d_x)}(p, \delta)$

Now look at the very last condition there:

for any $x \in B_{(X, d_x)}(p, \delta)$

What this actually states is that there must exist a $\delta > 0$ such that $B_{(X, d_x)}(p, \delta) \subseteq E$, because if $B_{(X, d_x)}(p, \delta) \not\subseteq E$, then there'd exist an $x \in B_{(X, d_x)}(p, \delta)$ such that $x \not \in E$, but then $f$ is not defined at $x$, since it lies outside of $E$, so the statement $$0 < d_X(x, p) < \delta \implies d_Y((f(x), q) < \epsilon$$ is meaningless because $f(x)$ is not even defined.

Thus since there must exist a $\delta > 0$ such that $B_{(X, d_x)}(p, \delta) \subseteq E$, it follows that $p$ is an interior point of $E$. So $p \in \operatorname{Int}_X(E)$ (which is an open set most certainly), that is $p$ is an element of the interior of $E$ with respect to the metric space $X$. Why is this the case? This is because above we've taken the open ball (open set) with respect to the metric space $X$.

That's all fine and good, but how does it affect differentiability you ask? Well differentiability depends on the limit of a function existing, we've seen above that the limit of a function doesn't exist (or make sense really) if we take the limit of a function at a non-interior point, so we can only take the limits of functions at interior points, which implies that we can only differentiate functions or talk about differentiation of functions defined on open sets. That is the main reason why we only define differentiability for open sets in $\mathbb{R}^n$. (Note how $\operatorname{Int}_X(E)$, translates to $\operatorname{Int}_{\mathbb{R}}\left((0, 1]\right)$ in my example above)

Consider the simple case of differentiability of a function between a subset of $A \subseteq \mathbb{R}$ and $\mathbb{R}$. Consider a function $f : A \to \mathbb{R}$, we define the derivative of $f$ at a point $a \in A$, as $$f'(a) = \lim_{t \to 0} \frac{f(a+t)-f(a)}{t}$$ provided the limit exists. Now for the time being suppose we don't include the added condition that $A$ must contain a neighborhood of $a$. Then note that there is no need for $f(a+t)$ to even be defined, so the limit doesn't make sense and we can't talk about the derivative of the function at $a$. So we really need the condition that $A$ must contain a neighborhood of $a$. I've given a more explicit example of this in the comments below.


Extra material

But there is another problem that I've yet to address, which is that the function $$\frac{f(a+t)-f(a)}{t}$$ isn't even given a domain in most textbooks, and you could say that for large $t$ it wouldn't even be defined if the domain of $f$ wasn't the whole of $\mathbb{R}$. The key realization to make is that we only need the function $\frac{f(a+t)-f(a)}{t}$ to be defined locally on a small neighborhood $\Theta_a$ containing $0$, and note very importantly that the neighborhood $\Theta_a$ containing $0$ depends directly on finding a neighborhood $\Omega$ containing $a$. Most textbooks don't stress the importance of this which is why I formalized the following definition to address these issues

My definition of the derivative of a function at a point: Let $U \subseteq \mathbb{R}$ be an open set. Let $f : U \to \mathbb{R}$ be any function. Pick $a \in U$ and choose $r > 0$ such that $B(a, r) \subseteq U$. Let $\Omega_a = B(a, r) = (a-r, a+r)$ and let $\Theta_{r_a} = B(0, r) \setminus \{0\}$ Define $\phi : \Theta_{r_a} \to \mathbb{R}$ by $$\phi(t) = \frac{f(a+t) - f(a)}{t}.$$ Then we define the derivative of $f$ at $a$ as $$f'(a) = \lim_{t \to 0} \phi(t)$$ provided that the limit exists and we say that $f$ is differentiable at $a$.

You can check based on the definitions I've given that this is indeed well-defined, and it is entirely equivalent to the definitions in Munkres' book and also pay attention to how the need for an open set is vital to this definition. But in practice I would never use this definition, this is just a way to check that everything your textbook authors are saying makes sense and is rigorous and that there is a reason that they've made the choices they have for a definition.

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  • $\begingroup$ About the last line of your answer; if $(0,1]$ open in $[0,1]$, then the set has to be equal to its interior, which is not what you are claiming, since you are still seeing the bigger space as $\mathbb{R}$, and not $[0,1]$ $\endgroup$ – onurcanbektas Aug 13 '18 at 11:12
  • $\begingroup$ Plus, $M$ was just an example, we do not have to see it as a subspace; we can define a different topology on it, and we would still be talking about the continuity of $f$. $\endgroup$ – onurcanbektas Aug 13 '18 at 11:20
  • $\begingroup$ @onurcanbektas Because that is how differentiation is usually defined. For example take a look at the definition of the derivative of a function $f : A \to \mathbb{R}$ where $A \subseteq \mathbb{R}$ (on page 41 of Munkres' Analysis on Manifolds). Suppose that $A = [0, 1]$ and assume $f$ is differentiable at $0$, then the limit $$f'(0) = \lim_{t\to 0} \frac{f(0+t)-f(0)}{t}$$ would be defined, but for any $t < 0$ we have that $f(0+t) = f(t)$ is not defined. So this limit can't even exist, so $f$ can't be differnetiable at $0$ $\endgroup$ – Perturbative Aug 13 '18 at 11:41
  • $\begingroup$ We need open sets of $\mathbb{R}$ in the above for our limits to behave nicely $\endgroup$ – Perturbative Aug 13 '18 at 11:42
  • $\begingroup$ so you are claiming that the reason why we do not have any definition of differentiability, or any generalisation of such a concept is because we haven't needed to, hence we basically did not so ? $\endgroup$ – onurcanbektas Aug 13 '18 at 11:57
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Not without extra structure. A function $f: X \to Y$ being a homeomorphism is the "nicest" kind of topological function: it means that the two spaces are the same topologically, and that $f$ functions more or less like the identity between the spaces, and thus should in particular be continuous. But there are plenty of homeomorphisms which are not differentiable; for instance, the function $f : [-2, 1] \to [-1, 1]$ defined by $$ f(x) = \begin{cases} x &\text{if $x \geq 0$}\\ 2x &\text{if $x \leq 0$} \end{cases} $$ is a homeomorphism which is not differentiable.

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  • $\begingroup$ For example, what kind of an extra structure would change the being differentiability of a single function depending on that extra structure ? $\endgroup$ – onurcanbektas Aug 13 '18 at 8:36
  • $\begingroup$ See my edit, please, because the answer (including yours) does not related to what I'm actually asking. $\endgroup$ – onurcanbektas Aug 13 '18 at 10:27

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