Open interval (set) end points and differentiability.

[kidsasd987]

When we talk about differentiability on a
Set X, the set has to be open.

And if a set X is open there exists epsilon> 0 where epsilon is in R.

Then if x is in X, y=x+ or - epsilon and y is also in X

But this contradicts to what i was taught in high school; end points are excluded in the open interval.

Could anyone clarify this for me?

Also, since epsilon is arbitrary number, if set it to be infinite then would X be R?
(Entire Real number set)

 

[Stephen Tashi]

And if a set X is open there exists epsilon> 0 where epsilon is in R.

Then if x is in X, y=x+ or - epsilon and y is also in X

That isn't a grammatically correct statement and it isn't the correct definition for "set X is open".

Also, since epsilon is arbitrary number, if set it to be infinite then would X be R?
(Entire Real number set)

A variable representing a single number can't be "set to be infinite" and it can't be set equal to a set of numbers.

 

[pwsnafu]

Let $X \subset \mathbb{R}$. We say $X$ is an open set if and only if for all $x \in X$ there exists $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon) \subset X$.

The order of the $\epsilon$ and $x$ is important: $\epsilon$ depends on $x$. This means it is not arbitrary.

 

[fresh_42]

When we talk about differentiability on a
Set X, the set has to be open...

Why? What about $f: [0,1] \rightarrow \mathbb{R}$ with $f(x) = x$ or $f(x) = \frac{1}{x}$? Why shouldn't we talk about differentiability here?

 

[Stephen Tashi]

Let $X \subset \mathbb{R}$. We say $X$ is an open set if and only if for all $x \in X$ there exists $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon) \subset X$.

The order of the $\epsilon$ and $x$ is important: $\epsilon$ depends on $x$. This means it is not arbitrary.

It's also clearer to say "for each $x \in X$" because we aren't insisting that there is a single $\epsilon$ that works for all $x$.

 

[micromass]

Why? What about $f: [0,1] \rightarrow \mathbb{R}$ with $f(x) = x$ or $f(x) = \frac{1}{x}$? Why shouldn't we talk about differentiability here?

It is not usually done in analysis books (at least not introductory ones). The question is how we should define differentiability at the boundary. You probably have quite a good idea how to do that, but it does complicate matters somewhat. Furthermore, differentiability at the boundary is almost never needed for elementary analysis results.

Also, when we generalize analysis to $\mathbb{R}^n$ it becomes even more difficult to describe differentiability at the boundary. It's not impossible to do it, but it takes some effort which is usually spent in other places.

In fact, I know of two good generalizations of differentiability at the boundary, and I'm not sure whether they even are equivalent. I should think a bit about this.

 

[fresh_42]

I basically wanted to animate the OP to think about why an open neighborhood is "needed", and to sharpen his argumentation, because "the set has to be open" isn't correct in this generality. A reflex, if you like. I thought it would help to understand differentials as linear approximations and what approximation really means in this context.
(I wonder if any derivation on any ring could be considered a differentiation ... with no metric in sight ...)

 

[micromass]

I basically wanted to animate the OP to think about why an open neighborhood is "needed", and to sharpen his argumentation, because "the set has to be open" isn't correct in this generality. A reflex, if you like. I thought it would help to understand differentials as linear approximations and what approximation really means in this context.
(I wonder if any derivation on any ring could be considered a differentiation ... with no metric in sight ...)

You mean in the sense of differential algebra? https://en.wikipedia.org/wiki/Differential_algebra

 

[fresh_42]

You mean in the sense of differential algebra? https://en.wikipedia.org/wiki/Differential_algebra

Yes. Beside I don't understand the constraint of finitely many derivations. What about Lie Algebras over infinite fields?

 

[kidsasd987]

You mean in the sense of differential algebra? https://en.wikipedia.org/wiki/Differential_algebra

Thanks. And I am sorry for the confusing statements and wording.

I sort of understand why we need an "open set" when we define differentiability on a set X. As far as I know, we'd need to define right or left differentiabilty if the set
is closed.

Let X⊂RX⊂RX \subset \mathbb{R}. We say XXX is an open set if and only if for all x∈Xx∈Xx \in X there exists ϵ>0ϵ>0\epsilon > 0 such that (x−ϵ,x+ϵ)⊂X(x−ϵ,x+ϵ)⊂X(x-\epsilon, x+\epsilon) \subset X.

But here I have two questions

1. I learned in high school that open intervals exclude end points. But in an open set X, it would include end points and its neighborhood.
Are they different in terms of definition?

2. If we consider the boundary points of X, intuitively half of (x-eps,x+eps) would include the interval outside the set X. Do we define (x-eps, x+eps) to be in X because epsilon is small?

 

[fresh_42]

1. I learned in high school that open intervals exclude end points.

That's correct.

But in an open set X, it would include end points and its neighborhood.

No. $(a,b) = \{x \in \mathbb{R} \, | \, a < x < b \}$ is an open set, $(a,b] = \{x \in \mathbb{R} \, | \, a < x \leq b \}$ is neither open nor closed, and $[a,b] = \{x \in \mathbb{R} \, | \, a \leq x \leq b \}$ is a closed interval.

2. If we consider the boundary points of X, intuitively half of (x-eps,x+eps) would include the interval outside the set X. Do we define (x-eps, x+eps) to be in X because epsilon is small?

What do you mean by "half of ... outside"?

We consider an open interval $(a,b)$ in $X = \mathbb{R}$, a point $x \in (a,b) \subset X$ and a neighborhood $(x-\epsilon,x+\epsilon) \subset (a,b)$ which is entirely within our open set / interval. We then speak of differentiability at (or in) $x$. The fact that $(a,b)$ is open and $x \in (a,b)$ that is $a < x < b$ guarantees us, that we can always find some neighborhood left and right of $x$ that is still in $(a,b)$ so we must not deal with any boundaries, where e.g. there is no limit from the left if it is the left end point. And we can get as close to $x$ as we like - from both sides.

 

[kidsasd987]

That's correct.

No. $(a,b) = \{x \in \mathbb{R} \, | \, a < x < b \}$ is an open set, $(a,b] = \{x \in \mathbb{R} \, | \, a < x \leq b \}$ is neither open nor closed, and $[a,b] = \{x \in \mathbb{R} \, | \, a \leq x \leq b \}$ is a closed interval.

What do you mean by "half of ... outside"?

We consider an open interval $(a,b)$ in $X = \mathbb{R}$, a point $x \in (a,b) \subset X$ and a neighborhood $(x-\epsilon,x+\epsilon) \subset (a,b)$ which is entirely within our open set / interval. We then speak of differentiability at (or in) $x$. The fact that $(a,b)$ is open and $x \in (a,b)$ that is $a < x < b$ guarantees us, that we can always find some neighborhood left and right of $x$ that is still in $(a,b)$ so we must not deal with any boundaries, where e.g. there is no limit from the left if it is the left end point. And we can get as close to $x$ as we like - from both sides.

Oh shoot. I see. Somehow I thought the open set would include the end points.. Thanks. It is so weird. where I got confused of this..