Proving R^n & Ø are the Only Subsets of R^n Open & Closed

[Math_Frank]

Homework Statement

The Premise:

Here One must prove that that R^n and Ø are the two subsets of R^n, which is both open and closed. You must that these are the only subsets of R^n which has this property!

Let $$X \subseteq \mathbb{R}^n$$ be a subset, which is both open and close, and here we must prove that if either X = R^n or X = \emptyset. Thusly $$X \neq R^n$$ and $$X \neq \emptyset.$$

Prove that this assumption leads to a contradiction:

Let $$Y = \mathbb{R}^n \setminus X$$ and show that Y is both open and closed and not empty!

Homework Equations

The Attempt at a Solution

This can only be the case if $$Y = \mathbb{R}^n \setminus \emptyset = \mathbb{R}^n$$ wouldn't it?

Cheers
Frank

 

[Dick]

If X and Y are both nonempty, pick x in X and y in Y. Now consider f(t)=x*(1-t)+y*t, L={f(t): t in [0,1]}. L is a line segment connecting x and y. Now think about the inverse image of X and Y intersected with L. You should be able to conclude that [0,1] can be split into two disjoint open sets whose union is [0,1]. Derive a contradiction from that.

 

[Math_Frank]

If X and Y are both nonempty, pick x in X and y in Y. Now consider f(t)=x*(1-t)+y*t, L={f(t): t in [0,1]}. L is a line segment connecting x and y. Now think about the inverse image of X and Y intersected with L. You should be able to conclude that [0,1] can be split into two disjoint open sets whose union is [0,1]. Derive a contradiction from that.

Hi Dick,

The leads me exactly to question

Let $$f: \mathbb{R}^n \rightarrow \mathbb{R}$$ be defined be

$$f(x) = \left[\[\begin{array}{cccc}1 \ \mathrm{for} \ x\in X, \\ 0 \ \mathrm{for} \ x\in Y, \\\end{array}$$

Then prove that f is continious for every $$x_0 \in X.$$

Proof:

If there exist a ball with center in every x0 in the neighbourhood X. Then every real function defined on that neighbourhood is continious.

Is that what is meant proving the continuity of that function? Hope I have understood this correctly:)

Best Regards
Frank.

 

[Dick]

Ok. Let's work with that. Yes, it's clear f(x) is continuous everywhere on R^n. Now as before take x in X and y in Y and write g(t)=x*(1-t)+y*t. g is a continuous map from [0,1] to R^n. So h(t)=f(g(t)) is a continuous map from [0,1] to [0,1]. h(0)=1 (since g(0)=x), h(1)=0 (since g(1)=y) and h takes on only the values 0 and 1. What's wrong with that picture?

 

[Math_Frank]

Ok. Let's work with that. Yes, it's clear f(x) is continuous everywhere on R^n. Now as before take x in X and y in Y and write g(t)=x*(1-t)+y*t. g is a continuous map from [0,1] to R^n. So h(t)=f(g(t)) is a continuous map from [0,1] to [0,1]. h(0)=1 (since g(0)=x), h(1)=0 (since g(1)=y) and h takes on only the values 0 and 1. What's wrong with that picture?

The X isn't open in Your case??

I know that the definition regarding the continious map (fnc between two topological space )can only be continious if the preimage is open. And since that we know that R^n is both open and closed, then f is continious for every x_o in X. Isn't that the answer for question 2?

Isn't it?

Sincerely Yours
Frank...

 

[HallsofIvy]

In order to prove that the only "clopen" (both closed and open) subsets of Rⁿ are Rⁿ itself and $\emptyset$ you are going to need the fact that Rⁿ is a connected set.

 

[Math_Frank]

In order to prove that the only "clopen" (both closed and open) subsets of Rⁿ are Rⁿ itself and $\emptyset$ you are going to need the fact that Rⁿ is a connected set.

Hello Hall and Dick,

2)

Proof:

R^n is connected which means that it cannot be partioned into two none-empty subsets, and if f is a continious map and therefore defined on the whole of R^n. Then f must also be continious for any x_0 on X, because is the pre-image of R^n, which is also open according to the definition.

3) Prove that f is continious for every y_o in Y. Since we know that the only version of Y which is both open and closed can be where Y = R^n \ (X = \emptyset), and since we know from 2) that any point on X is continious, then every y_0 point on Y must be continious as well??

Cheers.
Frank.

Have I understood it correctly??

 

[Dick]

The X isn't open in Your case??

I know that the definition regarding the continious map (fnc between two topological space )can only be continious if the preimage is open. And since that we know that R^n is both open and closed, then f is continious for every x_o in X. Isn't that the answer for question 2?

Isn't it?

Sincerely Yours
Frank...

Your replies are a little confusing. f(x) is continuous because x is in either X or Y and both X and Y are open. If it's in X, then it's constant and equal to one on a neighborhood of x. If it's in Y, then it's constant and equal to zero on a neighborhood of x. Think epsilon/delta.

 

[Math_Frank]

Your replies are a little confusing. f(x) is continuous because x is in either X or Y and both X and Y are open. If it's in X, then it's constant and equal to one on a neighborhood of x. If it's in Y, then it's constant and equal to zero on a neighborhood of x. Think epsilon/delta.

I am sorry, this CLopen set business is fairly new to me.

Then my professor tells me to construct a proof that f is continious for x_0 in X. Doesn't he simply want me to show if R^n is seen as Hallsoft writes a connected set, then all possible X_0 lies in the whole of R^n, making f continious on the whole neighbourhood R^n?

Best Regards
Frank.

 

[Dick]

Hello Hall and Dick,

2)

Proof:

R^n is connected which means that it cannot be partioned into two none-empty subsets, and if f is a continious map and therefore defined on the whole of R^n. Then f must also be continious for any x_0 on X, because is the pre-image of R^n, which is also open according to the definition.

3) Prove that f is continious for every y_o in Y. Since we know that the only version of Y which is both open and closed can be where Y = R^n \ (X = \emptyset), and since we know from 2) that any point on X is continious, then every y_0 point on Y must be continious as well??

Cheers.
Frank.

Have I understood it correctly??

I'm not sure I understand any of that, sorry. :(. If you KNOW R^n is connected then there is not much to prove. Have you proved that?

 

[Dick]

I am sorry, this CLopen set business is fairly new to me.

Then my professor tells me to construct a proof that f is continious for x_0 in X. Doesn't he simply want me to show if R^n is seen as Hallsoft writes a connected set, then all possible X_0 lies in the whole of R^n, making f continious on the whole neighbourhood R^n?

Best Regards
Frank.

You ALMOST stated a correct definition of continuous. f is continuous if the preimage of EVERY open set is open. Let's take this slowly. Take an open set in the reals. Suppose it contains both 0 and 1. Then what is it's preimage under f? Suppose it contains only 0. Then what's it's preimage? Etc.

 

[Math_Frank]

You ALMOST stated a correct definition of continuous. f is continuous if the preimage of EVERY open set is open. Let's take this slowly. Take an open set in the reals. Suppose it contains both 0 and 1. Then what is it's preimage under f? Suppose it contains only 0. Then what's it's preimage? Etc.

Hi Dick and Hallsoft,

Here is the proof of number (2)

Let $$f: \mathbb{R}^n \rightarrow \mathbb{R}$$ be defined as

$$f(x) = \left\{\begin{array}{ccc} 1 \ \ \mathrm{for} \ x \in X \\ 0 \ \ \mathrm{for} \ x \in Y \end{array}$$

Prove that f is continious in every $$x_0 \in X.$$

Proof

We know that X is either $$\mathbb{R}^n$$ or $$\emptyset$$. Assuming that X is a connected Space and that f is a two valued function on X. The trick is to show that f is constant. If $$\mathbb{R}^n = f^{-1}({0})$$ be the inverse image of the subsets $$\{0\}$$ and $$\{1\}$$. Assuming that {0} and {1} are open subsets of the metric space {0,1}, both $$\mathbb{R}^n$$ and $$\emptyset$$ are then open in X.

Assuming that any subset of X is nonempty, f takes both values 0 and 1, so f is not constant.

Also f is continious on X because the inverse image of every open subset of {0,1} is open in X.

Therefore the entire neighbourhood X is both connected and continuous for any element of X.

How does this sound?

Regards
Frank.

 

[Dick]

That sounds really unclear. DON'T assume R^n is connected. DON'T assume X=R^n or {}. That's basically what we are trying to prove, eventually. Don't even try to write the whole proof down, it gets all jumbled up when you do. ASSUME ONLY that X is open, Y is open, (X union Y)=R^n and (X intersect Y)={}. This a basically just a restatement of the assumption that X is clopen and labelling Y=(R^n/X). That's all. Now just two questions. i) what is f^(-1)({1})? ii) if A is an open subset of R that contains 1 but not 0, what is f^(-1)(A)?

 

[Math_Frank]

That sounds really unclear. DON'T assume R^n is connected. DON'T assume X=R^n or {}. That's basically what we are trying to prove, eventually. Don't even try to write the whole proof down, it gets all jumbled up when you do. ASSUME ONLY that X is open, Y is open, (X union Y)=R^n and (X intersect Y)={}. This a basically just a restatement of the assumption that X is clopen and labelling Y=(R^n/X). That's all. Now just two questions. i) what is f^(-1)({1})? ii) if A is an open subset of R that contains 1 but not 0, what is f^(-1)(A)?

Hi Dick,

Sorry that I am unclear,

f^(-1)({1}) that must be pre-image of X. And f^(-1)(A) is just be preimage of A?

Cheers,
Frank.

 

[Dick]

f^(-1)({1}) is not the preimage of X. It IS X, right? It's the PREIMAGE of {1}. You have to use words carefully in this game. Sure, f^(-1)(A) is the preimage of A, that's what preimage means. But what set that we are discussing IS the preimage of A? I've told you A contains 1 but not 0.

 

[HallsofIvy]

f^{-1}(A) is, by definition, the "preimage" of A- the set of all x such that f(x) is in A.

But f^{-1}({1}) is not the "preimag" of X, it is just the set of all x such that f(x)= 1.

 

[Math_Frank]

f^(-1)({1}) is not the preimage of X. It IS X, right? It's the PREIMAGE of {1}. You have to use words carefully in this game. Sure, f^(-1)(A) is the preimage of A, that's what preimage means. But what set that we are discussing IS the preimage of A? I've told you A contains 1 but not 0.

Sorry this rough terratory for me.

I am not asking for the proof, but could you direct me please to the which definition and theorem I need to build together to construct the needed proof?

I have a feeling its more than just connected space is connected?

Sincerely Frank.

 

[Dick]

Judging by the terminology you are using you already have all of the definitions and theorems you need to prove f is continuous (which is what you should concentrate on first). You just need to learn to put those concepts together grammatically. I'm going to ask you again. A contains 1 but doesn't contain 0. Look at the definition of f and tell me what f^(-1)(A) is. Think twice before you answer.

 

[Math_Frank]

Judging by the terminology you are using you already have all of the definitions and theorems you need to prove f is continuous (which is what you should concentrate on first). You just need to learn to put those concepts together grammatically. I'm going to ask you again. A contains 1 but doesn't contain 0. Look at the definition of f and tell me what f^(-1)(A) is. Think twice before you answer.

Since F is defined from R^n -> R, then as I understand it now f^(-1)(A) = 0 ?

I can connected space, continuity of a function(both locally and uniformally), but putting them together to prove the above.

I simply give up

Cheers,

Frank.

 

[HallsofIvy]

It is simply unbelievable to me that you are working with continuous functions and uniformly continuous functions but don't know what f^-1(A) means!

 

[HallsofIvy]

It is simply unbelievable to me that you are working with continuous functions and uniformly continuous functions but don't know what f^-1(A) means!

First of all, f^-1(A) can't be equal to 0, because f^-1(A) is a set, not a number! If you meant f^-1(A)= {0}, that would only be true if f(x) is NOT in A unless x= 0. If you mean f^-1(A)= { }, the empty set, that would only be true if f(x) is never in A.

 

[Math_Frank]

First of all, f^-1(A) can't be equal to 0, because f^-1(A) is a set, not a number! If you meant f^-1(A)= {0}, that would only be true if f(x) is NOT in A unless x= 0. If you mean f^-1(A)= { }, the empty set, that would only be true if f(x) is never in A.

You some textbook deal with different aspects differently and is the first time I have come across this kind of formulation. Believe it or not :(

Is easy for a guy like you who has been doing this for maybe thirty years, we newbee often to discover the solution straight away :(

Okay then the preimage of both R^n and emptyset aren't that then X because they both are subsets of X?

 

[Dick]

Since F is defined from R^n -> R, then as I understand it now f^(-1)(A) = 0 ?

I can connected space, continuity of a function(both locally and uniformally), but putting them together to prove the above.

I simply give up

Cheers,

Frank.

You are having some kind of a block here. And I guess it is that you haven't quite wrapped your head around the terminology. Ok, say an open set A in R contains 1 and does not contain 0. The preimage of A, f^(-1)(A) is the set of all x in R^n such that f(x) is in A. If you look at the definition of f, the only possible values are 1 or 0. Since A contains 1 and not 0, it must be that f(x)=1. Look again at the definition of f. That set is just X. X is open.

Here's where this is going. To prove f is continuous we need to prove that for every open set A in R, f^(-1)(A) is open. There are only four kinds of open sets. i) A contains 1 but not 0. ii) A contains 0 but not 1. iii) A contains both 0 and 1. iv) A contains neither 0 or 1. If you followed the preceding paragraph, you now know that f^(-1)(A)=X if A satisfies i). X is open. So f^(-1)(A) is open if A satisfies i). Suppose A satisfies ii)? What's f^(-1)(A)? Is it open? Ditto for iii) and iv). If you can answer that, yes, it is open for i) through iv) then f^(-1)(A) is open for ANY open set A. Hence, f is continuous. Whew!

 

[Dick]

Since F is defined from R^n -> R, then as I understand it now f^(-1)(A) = 0 ?

I can connected space, continuity of a function(both locally and uniformally), but putting them together to prove the above.

I simply give up

Cheers,

Frank.

You are having some kind of a block here. And I guess it is that you haven't quite wrapped your head around the terminology. Ok, say an open set A in R contains 1 and does not contain 0. The preimage of A, f^(-1)(A) is the set of all x in R^n such that f(x) is in A. If you look at the definition of f, the only possible values are 1 or 0. Since A contains 1 and not 0, it must be that f(x)=1. Look again at the definition of f. That set is just X. X is open.

Here's where this is going. To prove f is continuous we need to prove that for every open set A in R, f^(-1)(A) is open. There are only four kinds of open sets. i) A contains 1 but not 0. ii) A contains 0 but not 1. iii) A contains both 0 and 1. iv) A contains neither 0 nor 1. If you followed the preceding paragraph, you now know that f^(-1)(A)=X if A satisfies i). X is open. So f^(-1)(A) is open if A satisfies i). Suppose A satisfies ii)? What's f^(-1)(A)? Is it open? Ditto for iii) and iv). If you can answer that, yes, it is open for i) through iv) then f^(-1)(A) is open for ANY open set A. Hence, f is continuous. Whew!

 

[Math_Frank]

You are having some kind of a block here. And I guess it is that you haven't quite wrapped your head around the terminology. Ok, say an open set A in R contains 1 and does not contain 0. The preimage of A, f^(-1)(A) is the set of all x in R^n such that f(x) is in A. If you look at the definition of f, the only possible values are 1 or 0. Since A contains 1 and not 0, it must be that f(x)=1. Look again at the definition of f. That set is just X. X is open.

Here's where this is going. To prove f is continuous we need to prove that for every open set A in R, f^(-1)(A) is open. There are only four kinds of open sets. i) A contains 1 but not 0. ii) A contains 0 but not 1. iii) A contains both 0 and 1. iv) A contains neither 0 nor 1. If you followed the preceding paragraph, you now know that f^(-1)(A)=X if A satisfies i). X is open. So f^(-1)(A) is open if A satisfies i). Suppose A satisfies ii)? What's f^(-1)(A)? Is it open? Ditto for iii) and iv). If you can answer that, yes, it is open for i) through iv) then f^(-1)(A) is open for ANY open set A. Hence, f is continuous. Whew!

Oh Thank You Thank You. That was the templete that I was looking :) I hope I haven't insulted Hall. You are right the terminology of this hard for a newbie like me.

 

[Dick]

Great. If you think you REALLY understand that then go back and read some of your previous answers. If you CAN'T understand them, and you can sympathize with Hall's impatience, then you really have learned something.

 

[Math_Frank]

Great. If you think you REALLY understand that then go back and read some of your previous answers. If you CAN'T understand them, and you can sympathize with Hall's impatience, then you really have learned something.

Thank You Dick,

You are a good teacher:) I will go back read the answers and then add them together to the desired proof.

I will post it later.

Sincerely
Frank