Welcome to our community

Be a part of something great, join today!

Frontiers (Boundaries) in the Plane ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,881
Hobart, Tasmania
I am reading Stephen Willard: General Topology ... ... and am studying Chapter 2: Topological Spaces and am currently focused on Section 3: Fundamental Concepts ... ...

I need help in order to prove the assertion in Willard Exercise 3B ...


The assertion in Willard Exercise 3B reads as follows:




Willard ... Exercise 3B .png



I am assuming that Willard is assuming the usual topology and metric in \(\displaystyle \mathbb{R}^2\) ... and so a set \(\displaystyle E\) is open iff for each \(\displaystyle x \in E\) there is an \(\displaystyle \epsilon\)-disk (open ball) \(\displaystyle U(x, \epsilon)\) about x contained in \(\displaystyle E\) ...

Can someone please rigorously demonstrate the truth of assertion 3B ...

Help will be much appreciated ...

Peter


===============================================================================================================

Readers of the above post may be helped by access to Willard's definition of an \(\displaystyle \epsilon\)-disk (open ball) and an open set in a metric space so I am providing access to the same ... as follows:


Willlard ... e-disk and open set in a metric space .png



Readers of the above post may also be helped by access to Willard's definition of frontier (boundary) and some basic results regarding the frontier of a set .... so I am providing access to the same ... as follows: ...



Willard ... Frontier or boundary .png


Readers of the above post may also be helped by access to Willard's definition of closure and some basic results regarding the closure of a set .... so I am providing access to the same ... as follows: ...


Willard - Defn 3.5, Lemma 3..6 and Theorem 3.7 .png



Readers of the above post may also be helped by access to Willard's definition of interior and some basic results regarding theinterior of a set .... so I am providing access to the same ... as follows: ...


Willard - Interior ... Defn 3.9, Lemma 3.10 and Theorem 3.11 .png



Hope that helps ...

Peter
 

Attachments

Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,649
Leeds, UK
I am reading Stephen Willard: General Topology ... ... and am studying Chapter 2: Topological Spaces and am currently focused on Section 3: Fundamental Concepts ... ...

I need help in order to prove the assertion in Willard Exercise 3B ...


The assertion in Willard Exercise 3B reads as follows:




View attachment 10017



I am assuming that Willard is assuming the usual topology and metric in \(\displaystyle \mathbb{R}^2\) ... and so a set \(\displaystyle E\) is open iff for each \(\displaystyle x \in E\) there is an \(\displaystyle \epsilon\)-disk (open ball) \(\displaystyle U(x, \epsilon)\) about x contained in \(\displaystyle E\) ...

Can someone please rigorously demonstrate the truth of assertion 3B ...
To start with a bit of notation, I'll denote the boundary of $E$ (I prefer to call it the boundary rather than the frontier) by $\partial E$.

As far as I can see, this Exercise 3B is not easy. To see why, consider the example where $E$ is the closed unit disk in $\Bbb{R}^2$: $E = \{(x,y) \in \Bbb{R}^2: \|(x,y)\| \leqslant 1\}.$ Its boundary is the unit circle, and its interior is the open unit disc $\{(x,y) \in \Bbb{R}^2: \|(x,y)\| < 1\}.$ Can you think of a set $X\subset \Bbb{R}^2$ such that $E = \partial X$? It took me a while to realise that a possible answer would be to take $X$ to be the set of all elements in $E$ whose coordinates are rational: $X = \{(x,y) \in \Bbb{Q}^2: \|(x,y)\| \leqslant 1\}.$ Then the closure of $X$ is equal to the whole of $E$.

If you can see why that example works, I think it may give a clue to finding the general solution to the problem. If $E$ is a closed set in $\Bbb{R}^2$ then $E$ consists of its boundary together with its interior: $E = E^\circ \cup \partial E$. There is a theorem saying that an open set in $\Bbb{R}^n$ contains a countable dense subset. So let $Y$ be a countable dense subset of $E^\circ$, and let $X = Y \cup \partial E$. Then it should be true that $E = \overline{X}$.

I hope I am not making this problem too complicated. Maybe someone can come up with a simpler approach?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,881
Hobart, Tasmania
Thanks for your analysis Opalg ...

Thanks also for a rather interesting and I must say stunning example ...

I am still puzzling over this exercise...

I note in passing that Willard doesn’t cover countable dense subsets until much later in his book ... so like you I hope someone can come up with a simpler approach ...

Thanks again for the analysis and the example ...

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,881
Hobart, Tasmania
To start with a bit of notation, I'll denote the boundary of $E$ (I prefer to call it the boundary rather than the frontier) by $\partial E$.

As far as I can see, this Exercise 3B is not easy. To see why, consider the example where $E$ is the closed unit disk in $\Bbb{R}^2$: $E = \{(x,y) \in \Bbb{R}^2: \|(x,y)\| \leqslant 1\}.$ Its boundary is the unit circle, and its interior is the open unit disc $\{(x,y) \in \Bbb{R}^2: \|(x,y)\| < 1\}.$ Can you think of a set $X\subset \Bbb{R}^2$ such that $E = \partial X$? It took me a while to realise that a possible answer would be to take $X$ to be the set of all elements in $E$ whose coordinates are rational: $X = \{(x,y) \in \Bbb{Q}^2: \|(x,y)\| \leqslant 1\}.$ Then the closure of $X$ is equal to the whole of $E$.

If you can see why that example works, I think it may give a clue to finding the general solution to the problem. If $E$ is a closed set in $\Bbb{R}^2$ then $E$ consists of its boundary together with its interior: $E = E^\circ \cup \partial E$. There is a theorem saying that an open set in $\Bbb{R}^n$ contains a countable dense subset. So let $Y$ be a countable dense subset of $E^\circ$, and let $X = Y \cup \partial E$. Then it should be true that $E = \overline{X}$.

I hope I am not making this problem too complicated. Maybe someone can come up with a simpler approach?
The more I reflect on your post the more I think you have solved the exercise as simply as is possible...

‘Thanks again

Peter