Is Y Union A Connected When X Minus Y Splits into Disjoint Subsets A and B?

[PsychonautQQ]

Homework Statement

Let Y be a subspace of X and let both X and Y be connected. If X-Y=AUB where the intersection of A and B is empty, show that YUA is connected.

Homework Equations

The Attempt at a Solution

Say YUA = CUD where C and D are disjoint.

Let C_y be the intersection of Y with C and D_y be the intersection of D with Y.
Since A and Y have an empty intersection, Y=D_y U C_y, but since Y is connected this means that either D_y or C_y is empty, or in other words that Y is completely contained in either C or D.

Am I on the right track here? I am quite stuck now

 

[andrewkirk]

Are you sure there isn't some information missing? In particular, I wonder if we were supposed to assume that A and B are open sets.

If not, the statement 'X-Y=AUB where the intersection of A and B is empty' tells us nothing, because for any set A, we get a disjoint union of that form simply by defining $B=X-Y-A$.

 

[PsychonautQQ]

Yes, I'm sorry, A and B are open sets

 

[PsychonautQQ]

So wlog let us assume that C is contained in Y,, if we can show that A must also be contained in C then YUA is connected because D would be empty. I have not yet used the information that X is connected. Perhaps we can a assume that the intersection of A and D is nontrivial and show that this leads to a contradiction by connectedness of X?

 

[WWGD]

Homework Statement

Let Y be a subspace of X and let both X and Y be connected. If X-Y=AUB where the intersection of A and B is empty, show that YUA is connected.

Homework Equations

The Attempt at a Solution

Say YUA = CUD where C and D are disjoint.

Let C_y be the intersection of Y with C and D_y be the intersection of D with Y.
Since A and Y have an empty intersection, Y=D_y U C_y, but since Y is connected this means that either D_y or C_y is empty, or in other words that Y is completely contained in either C or D.

Am I on the right track here? I am quite stuck now

You mean $Y \cup A$ is connected as a subspace of X ?

 

[PsychonautQQ]

Yeah

 

[PsychonautQQ]

Okay so let me go over where I'm at, if somebody has any advice that'd be swell :D.

Recapping what we know:
Y is a subspace of X, both Y and X are connected.
X-Y=AUB where AiB = empty (i = intersection) and A and B are open sets in X-Y.
WWTS that YUA and YUB are both connected as subspaces of X

Then I supposed en route of a contradiction that YUA = DUC where DiC = empty and D and C are open in YUA.
YiA is empty because A is contained in the complement of Y.
D_y = DiY
C_y = CiY
So Y = D_y U C_y, but since Y is connected that must mean that either D_y or C_y is empty
WLOG let us say that D_y is empty and so Y < C (Y is contained in C).

So if I can somehow show that A is also contained in C, aka t hat DiA = Empty, then D is empty and hence YUA is connected. I haven't yet used the part our assumption that X is connected; So I'm assuming that DiA is not empty and then looking for a reason that this contradicts the connectedness of X.

Is this making sense? I hope so. Here is some more stuff:

D = D' i (YUA)
C = C' i (YUA)
where D' and C' are open in X. We know that such a D' and C' must exist by the definition of D being an open set in YUA where YUA is a subspace of X.

Also,
A = A' i (X-Y)
B = B' i (X-Y)
where A' and B' are open in X. We know that such a A' and B' must exist because X-Y is a subspace of of X and A and B are open sets in X-Y.

Anyone have any advice?

 

[PsychonautQQ]

Say that Y is a subspace of X and both X and Y are connected. If X-Y is separable (X-Y = AUB where AiB = empty and A and B are open in X-Y) then show that YUA and YUB are connected.

Okay so let me go over where I'm at, if somebody has any advice that'd be swell :D.

Recapping what we know:
Y is a subspace of X, both Y and X are connected.
X-Y=AUB where AiB = empty (i = intersection) and A and B are open sets in X-Y.
WWTS that YUA and YUB are both connected as subspaces of X

Then I supposed en route of a contradiction that YUA = DUC where DiC = empty and D and C are open in YUA.
YiA is empty because A is contained in the complement of Y.
D_y = DiY
C_y = CiY
So Y = D_y U C_y, but since Y is connected that must mean that either D_y or C_y is empty
WLOG let us say that D_y is empty and so Y < C (Y is contained in C).

So if I can somehow show that A is also contained in C, aka t hat DiA = Empty, then D is empty and hence YUA is connected. I haven't yet used the part our assumption that X is connected; So I'm assuming that DiA is not empty and then looking for a reason that this contradicts the connectedness of X.

Is this making sense? I hope so. Here is some more stuff:

D = D' i (YUA)
C = C' i (YUA)
where D' and C' are open in X. We know that such a D' and C' must exist by the definition of D being an open set in YUA where YUA is a subspace of X.

Also,
A = A' i (X-Y)
B = B' i (X-Y)
where A' and B' are open in X. We know that such a A' and B' must exist because X-Y is a subspace of of X and A and B are open sets in X-Y.

Anyone have any advice?

 

[fresh_42]

I have inserted the LaTeX commands to increase readability (and hopefully without errors). It is still hard to read as you deal with a total of $10$ - in words TEN - different sets: $X,Y,A,B,C,D,C',D',C_y,D_y\,$, which disguises your argument quite well. One needs a machete to go through this ...

Say that $Y$ is a subspace of $X$ and both $X$ and $Y$ are connected. If $X-Y$ is separable $(\,X-Y = A\cup B$ where $A \cap B = \emptyset$ and $A$ and $B$ are open in $X-Y\,)$ then show that $Y \cup A$ and $Y \cup B$ are connected.

Okay so let me go over where I'm at, if somebody has any advice that'd be swell :D.

Recapping what we know:
$Y$ is a subspace of $X$, both $Y$ and $X$ are connected.
$X-Y=A \cup B$ where $A\cap B = \emptyset$ and $A$ and $B$ are open sets in $X-Y$.
WWTS that $Y \cup A$ and $Y \cup B$ are both connected as subspaces of $X$

Then I supposed en route of a contradiction that $Y \cup A$ = $D \cup C$ where $D \cap C = \emptyset$ and $D$ and $C$ are open in $Y \cup A$.
$Y \cap A$ is empty because $A$ is contained in the complement of $Y$.
$D_y = D \cap Y$
$C_y = C \cap Y$
So $Y = D_y \cup C_y$, but since $Y$ is connected that must mean that either $D_y$ or $C_y$ is empty
WLOG let us say that $D_y$ is empty and so $Y \subseteq C$.

So if I can somehow show that $A$ is also contained in $C$, aka that $D \cap A = \emptyset$, then $D$ is empty and hence $Y \cup A$ is connected. I haven't yet used the part our assumption that $X$ is connected; So I'm assuming that $D \cap A \neq \emptyset$ and then looking for a reason that this contradicts the connectedness of $X$.

Is this making sense? I hope so. Here is some more stuff:

$D = D' \cap (Y \cup A)$
$C = C' \cap (Y \cup A)$
where $D'$ and $C'$ are open in $X$. We know that such a $D'$ and $C'$ must exist by the definition of $D$ being an open set in $Y \cup A$ where $Y \cup A \subseteq X$.

Also,
$A = A' \cap (X-Y)$
$B = B' \cap (X-Y)$
where $A'$ and $B'$ are open in $X$. We know that such a $A'$ and $B'$ must exist because $X-Y$ is a subspace of of $X$ and $A$ and $B$ are open sets in $X-Y$.

 

[Dick]

Okay so let me go over where I'm at, if somebody has any advice that'd be swell :D.

Recapping what we know:
Y is a subspace of X, both Y and X are connected.
X-Y=AUB where AiB = empty (i = intersection) and A and B are open sets in X-Y.
WWTS that YUA and YUB are both connected as subspaces of X

Then I supposed en route of a contradiction that YUA = DUC where DiC = empty and D and C are open in YUA.
YiA is empty because A is contained in the complement of Y.
D_y = DiY
C_y = CiY
So Y = D_y U C_y, but since Y is connected that must mean that either D_y or C_y is empty
WLOG let us say that D_y is empty and so Y < C (Y is contained in C).

So if I can somehow show that A is also contained in C, aka t hat DiA = Empty, then D is empty and hence YUA is connected. I haven't yet used the part our assumption that X is connected; So I'm assuming that DiA is not empty and then looking for a reason that this contradicts the connectedness of X.

Is this making sense? I hope so. Here is some more stuff:

D = D' i (YUA)
C = C' i (YUA)
where D' and C' are open in X. We know that such a D' and C' must exist by the definition of D being an open set in YUA where YUA is a subspace of X.

Also,
A = A' i (X-Y)
B = B' i (X-Y)
where A' and B' are open in X. We know that such a A' and B' must exist because X-Y is a subspace of of X and A and B are open sets in X-Y.

Anyone have any advice?

If $Y \subset C$ it must be that $A \cap D$ is nonempty. Try find open sets that separate $A \cap D$ from the rest of $X$ giving you a contradiction to $X$ being connected.

 

[jim mcnamara]

@fresh_42 Your post looks like your work, not the OP's I find this thread confusing enough without trying to get who wrote what.
Is there some way you could set off your translation like maybe HTML quote tags or something?

@PsychonautQQ -
You have a LOT of objects in your sketch. Fresh is right - using latex would help us poor old fogies read your post. PF has a nice page on latex composing for posts.

 

[fresh_42]

@fresh_42 Your post looks like your work, not the OP's I find this thread confusing enough without trying to get who wrote what.

Corrected. I only thought it would be easier to read outside the quotation tags. Sorry.

using latex would help us poor old fogies read your post

@PsychonautQQ It is rather simple: Already the usage of $\text{$ math symbol $}$ around your set names would have helped a lot. And whether you type U or $\text{ \cup }$ and i or $\text{ \cap }$ isn't so much trouble. Or just use the symbols offered behind the $\Sigma$ symbol in our post editor, where the most common symbols could be added the same way as smileys can be used.

 

[PsychonautQQ]

Ugh okay guys sorry I've put off learning latex too long I see now it's criticalness

 

[PsychonautQQ]

If $Y \subset C$ it must be that $A \cap D$ is nonempty. Try find open sets that separate $A \cap D$ from the rest of $X$ giving you a contradiction to $X$ being connected.

LaTex coming soon... So since Y is contained in C it must be that AiD is nonempty other YUA = CUD wouldn't be a separation of YUA since D would be empty; cool I'm following.

Now you say I should find sets that separate AiD from the rest of X, because then I would have found a separation in X which is a contradiction because X is connected by hypothesis. But does the fact that a separation must be formed by disjoint OPEN sets? I suppose I could show that AiD will be open, but how do I know the other sets that form the separation will be as well?

 

[Dick]

LaTex coming soon... So since Y is contained in C it must be that AiD is nonempty other YUA = CUD wouldn't be a separation of YUA since D would be empty; cool I'm following.

Now you say I should find sets that separate AiD from the rest of X, because then I would have found a separation in X which is a contradiction because X is connected by hypothesis. But does the fact that a separation must be formed by disjoint OPEN sets? I suppose I could show that AiD will be open, but how do I know the other sets that form the separation will be as well?

$C$ and $D$ are disjoint open sets that contain all of $X$ except for $B$. Replace $D$ with $A \cap D$, that's an open set , right? Does it intersect $B$?

 

[PsychonautQQ]

But C and D are disjoint sets that are open in YUA, not necessarily open in X. That's why I brought in the D' and C' that would be open in X and their intersection with YUA would be D and C, respectively. But then the thing is I don't know if D' and C' are exactly what I need them to be.

Also, B is open in X-Y not necessarily in X.

Right?

 

[fresh_42]

I have a question about your technical terms.
I know separable in a topological context as something very different.
What you might mean is called separated here, so there might be a variation due to translation. However, two sets $A,B$ are separated, if $\overline{A} \cap B = A \cap \overline{B} = \emptyset$, which is different from what you wrote.
As a connected set $X$ to my knowledge is defined as a set that cannot be written as a union of two non-empty separated sets, this tiny differences might play a role.

I just have started to understand what you wrote, so it might be of no harm, but I don't want to do it twice.

 

[Dick]

But C and D are disjoint sets that are open in YUA, not necessarily open in X. That's why I brought in the D' and C' that would be open in X and their intersection with YUA would be D and C, respectively. But then the thing is I don't know if D' and C' are exactly what I need them to be.

Also, B is open in X-Y not necessarily in X.

Right?

Right, but you are letting yourself get tangled up in which topology a set is open in, which is obscuring the point. This problem is about connectivity, not openness. Let's state the problem in its most general terms. Let $X$ and $Y$ be connected subsets of some set $U$. So the only open sets we will talk about are open in $U$. Suppose $X-Y=A \cup B$ and there are open sets $N$ and $M$ such that $A \subseteq N$ and $B \subseteq M$ and $N \cap M=\phi$. No need to assume $A$ and $B$ are themselves open, for example. Now show $Y \cup A$ is connected.

 

[fresh_42]

Two threads have been merged.

 

[andrewkirk]

I have a solution for this problem (at least I think it's one. Possibly it has flaws that others may spot).
I'm uncertain as to whether this problem is homework though, since it was posted in a non-homework as well as a homework forum. If it is homework, I'd better not post the solution, only hints.

It's easier to post the solution though, because I've already written it (for my own entertainment), and judging an appropriate level for a hint is quite tricky for a complex problem like this.

So before going any further,
@PsychonautQQ Is this homework, or just something you're doing for fun?

 

[Dick]

It's easier to post the solution though, because I've already written it (for my own entertainment), and judging an appropriate level for a hint is quite tricky for a complex problem like this.

If the problem is getting 'complex' that's probably a hint you've taken a wrong turn. The OP was almost there.

 

[PsychonautQQ]

I have a solution for this problem (at least I think it's one. Possibly it has flaws that others may spot).
I'm uncertain as to whether this problem is homework though, since it was posted in a non-homework as well as a homework forum. If it is homework, I'd better not post the solution, only hints.

It's easier to post the solution though, because I've already written it (for my own entertainment), and judging an appropriate level for a hint is quite tricky for a complex problem like this.

So before going any further,
@PsychonautQQ Is this homework, or just something you're doing for fun?

Nope, it's not homework, just doing a self-study in topology before my MGRE in... 9ish months haha. I suspected that this problem was trickier then it appears on the surface, especially when we are using Munkre's definition of connectedness where the two disjoint subsets must be open in whatever topology you are working in.

 

[PsychonautQQ]

I have a question about your technical terms.
I know separable in a topological context as something very different.
What you might mean is called separated here, so there might be a variation due to translation. However, two sets $A,B$ are separated, if $\overline{A} \cap B = A \cap \overline{B} = \emptyset$, which is different from what you wrote.
As a connected set $X$ to my knowledge is defined as a set that cannot be written as a union of two non-empty separated sets, this tiny differences might play a role.

I just have started to understand what you wrote, so it might be of no harm, but I don't want to do it twice.

The way Munkre's 'First Course in Topology' describes connected sets as what you said but with the added condition that the two disjoint subsets whose union is the whole set must be open sets in whatever topology you're working in. Now if a subspace of the original space is not connected then you will be able to find two subsets contained in the subspace that are open in the subspace topology whose union is the subspace and neither contains a limit point of the other; the fact that neither contains a limit point of the other will be a consequence of the way Munkre's is defining connectedness.

 

[PsychonautQQ]

Right, but you are letting yourself get tangled up in which topology a set is open in, which is obscuring the point. This problem is about connectivity, not openness. Let's state the problem in its most general terms. Let $X$ and $Y$ be connected subsets of some set $U$. So the only open sets we will talk about are open in $U$. Suppose $X-Y=A \cup B$ and there are open sets $N$ and $M$ such that $A \subseteq N$ and $B \subseteq M$ and $N \cap M=\phi$. No need to assume $A$ and $B$ are themselves open, for example. Now show $Y \cup A$ is connected.

The way Munkre's 'First course in topology' (first edition) defines connectedness is that you won't be able to find two open disjoint subsets whose union is the whole set. So the definition of connectedness that I'm using involves openness.

 

[Dick]

The way Munkre's 'First course in topology' (first edition) defines connectedness is that you won't be able to find two open disjoint subsets whose union is the whole set. So the definition of connectedness that I'm using involves openness.

Right. But $A$ and $B$ themselves don't have to be open. They just have to be separated by open sets, $M$ and $N$ in my restatement.

 

[PsychonautQQ]

Right, but you are letting yourself get tangled up in which topology a set is open in, which is obscuring the point. This problem is about connectivity, not openness. Let's state the problem in its most general terms. Let $X$ and $Y$ be connected subsets of some set $U$. So the only open sets we will talk about are open in $U$. Suppose $X-Y=A \cup B$ and there are open sets $N$ and $M$ such that $A \subseteq N$ and $B \subseteq M$ and $N \cap M=\phi$. No need to assume $A$ and $B$ are themselves open, for example. Now show $Y \cup A$ is connected.

How do we know that $N∩M=∅$?

 

[Dick]

How do we know that $N∩M=∅$?

It's what we are assuming as a premise. Your original statement of the problem corresponds the the special case $A=N$ and $B=M$.

 

[PsychonautQQ]

It's what we are assuming as a premise. Your original statement of the problem corresponds the the special case $A=N$ and $B=M$.

In this special case you are referring to, would X=U?

p.s.
Okay, cool, thanks for being so patient with me. Are you sure you arn't oversimplifying things? I mean you and AndrewKirk have both shown yourselves to be very smart in the past and you guys have differing opinions right now so that's unnerving to me, but i'll contemplate your post until my feeble brain starts clicking.

 

[Dick]

In this special case you are referring to, would X=U?

p.s.
Okay, cool, thanks for being so patient with me. Are you sure you arn't oversimplifying things? I mean you and AndrewKirk have both shown yourselves to be very smart in the past and you guys have differing opinions right now so that's unnerving to me, but i'll contemplate your post until my feeble brain starts clicking.

Sure, and $Y$ is a subset of $X$. And none of those things is necessary. And, I don't think we have a difference of opinion. It's always possible to solve a simple problem correctly in a complex way. But see if you can beat AndrewKirk to it. The way I stated it in post 18 is the most general way to state the problem. There is no assumption about $X, Y, A, B$ being open or closed. Remember the idea is to use your $C, D$ separation of $A \cup Y$ to show there is a separation of $X$. It's not so hard. You can define it completely in terms of the open sets $C, D, M, N$, without introducing a bunch of other sets. Really!

 

[andrewkirk]

Nope, it's not homework, just doing a self-study in topology

Good for you. Self-study is brilliant way to spend one's spare time.

Here's my proof. I suspect it can be streamlined a bit, as a few steps get repeated, but I thought I'd just post it as is, just to get it out there.

The proof works mostly in terms of limit points rather than open sets, as the open-set approach was getting too confusing by generating so many open sets. We use the notion that if two sets are 'connected' they share a limit point. Since X is connected and X-Y isn't, and A is disconnected from B, every component of A must be connected to Y.

IIRC the first part of the proof mirrors yours.

Here goes:
------------------------------------------------------------------------------------------------------------------------
We will use three lemmas, shown after the end of this proof. One is from Munkres.

Assume there is a separation $C,D$ of $Y\cup A$.

Let $\mathscr S$ be the collection of components of $A$. Then $Y$ must have a non-empty intersection with at least one of $C,D$. Assume wlog it is $C$. Then $Y\subseteq C$ because if not, $(Y\cap C),(Y\cap D)$ is a separation of $Y$, which cannot exist because $Y$ is connected.

Consider a component $A'\in\mathscr S$
Then $A'$ must have a non-empty intersection with at least one of $C,D$. Assume it is $C$. Then $A'\subseteq C$ because if not, $(A'\cap C),(A'\cap D)$ is a separation of $A'$, which cannot exist because $A'$ is connected. Similarly, if $A'\cap D\neq\emptyset$ then $A'\subseteq D$.

So every component of $A$ lies entirely in either $C$ or $D$, and we can write

$$C=Y\cup\bigcup_{\substack{A'\in\mathscr S\\A'\subseteq C}} A'
\quad\quad \quad\quad
D=\bigcup_{\substack{A'\in\mathscr S\\A'\subseteq D}} A'$$

Hence $D\subseteq A$.

Since $C,D$ is a separation of $Y\cup A$, by Lemma 1.1 neither of $C$ or $D$ contains a limit point of the other, in the topology of $Y\cup A$. Using Lemma 3, we conclude that neither of $C$ or $D$ contains a limit point of the other in the topology of $X$.

We also know $A,B$ is a separation of $X-Y$ so, by Lemma 1.1, neither contains a limit point of the other in the topology of $X-Y$. Hence, since $D$ is a subset of $A$, it cannot contain any limit points of $B$ in the $X-Y$ topology. And since the set of limit points of $D$ is a subset of the set of limit points of $A$, and $B$ contains no limit points of $A$, $B$ also contains no limit points of $D$. So neither of $D$ and $B$ contains a limit point of the other in the topology of $X-Y$ and, by Lemma 3, neither contains a limit point of the other in the topology of $X$.

Since neither $B$ nor $C$ contains any limit points of $D$, it follows that $B\cup C$ contains no limit points of $D$.

Also $D$ contains no limit points of either $B$ or $C$. We now claim that $D$ contains no limit points of $B\cup C$. To prove that, assume the contrary , that $D$ contains a limit point $x$ of $B\cup C$. That cannot be a limit point of $B$ or $C$ so there must be open sets $U_B,U_C$, both containing $x$, such that $U_B\cap B= U_C\cap C= \emptyset$. But then $U_B\cap U_C$ is an open set containing $x$, which intersects neither $B$ nor $C$ and hence does not intersect $B\cup C$. So $x$ is not a limit point of $B\cup C$ and we have a contradiction. Hence $D$ contains no limit points of $B\cup C$.

So neither of $D$ and $B\cup C$ contains a limit point of the other. So both contain all their limit points. So both are closed in $X$ and, since each is the complement of the other, both are open. Hence $D,B\cup C$ is a separation of $X$, which contradicts the connectedness of $X$.

Hence the assumption that there was a separation $C,D$ of $Y\cup A$ must be false. $Y\cup A$ is connected.

-------------------------------------------------------

Munkres Lemma 1.1

If $V,W$ form a separation of $S$ then neither of $V,W$ contains any limit points of the other.

Lemma 2

If $V,W$ are subspaces of $S$ and $V$ contains a limit point of $W$ in the $S$ topology then $V$ contains a limit point of $W$ in the $V\cup W$ subspace topology.

Proof: Let $x$ be such a limit point. Then for every open set $H$ open in $S$ that contains $x$, there exists $y\in W\cap H,\ y\neq x$. Consider an arbitrary set $H'$ that is open in $V\cup W$ and contains $x$. This is $H\cap (V\cup W)$ for some $H$ open in $S$. Then there exists $y\in W\cap H,\ y\neq x$. But since $H'=H\cap (V\cup W)=(H\cap V)\cup(H\cap W)$, we have $y\in H'\cap W$. QED (for Lemma).

Lemma 3

If $V,W$ are subspaces of $S$, and $V$ contains no limit points of $W$ in the $V\cup W$ subspace topology then $V$ contains no limit points of $W$ in the $S$ topology.

This is simply the contrapositive of Lemma 2.

 

[Dick]

Let $X$ and $Y$ be connected subsets of some set $U$. Suppose $X-Y=A \cup B$ and there are open sets $N$ and $M$ such that $A \subseteq N$ and $B \subseteq M$ and $N \cap M=\phi$. Now show $Y \cup A$ is connected.

Assume $C$ and $D$ separate $Y \cup A$ such that $Y \subseteq C$. Then clearly $A \cap D$ is nonempty. Then $D \cap N$ (this part contains $A \cap D$) and $C \cup M$ (this contains the rest of $X$) separate $X$. They are disjoint because $C \cap D=\phi$ and $N \cap M=\phi$. This contradicts $X$ being connected hence $Y \cup A$ is connected. QED.

I don't know why this is so elusive...

 

[andrewkirk]

I don't know why this is so elusive...

It's not elusive. It's just a different problem from the one in the OP.

 

[Dick]

It's not elusive. It's just a different problem from the one in the OP.

In what way?? As I've noted before the original problem is the special case $A=N$, $B=M$, $U=X$. The proof of the special case is the same as the proof of the general.

 

[PsychonautQQ]

I am also curious as to why what dick did was a diff problem, it seems legit to me. Still a little worried about how we know A∩D and C∪B are open in X though.

 

[Dick]

I am also curious as to why what dick did was a diff problem, it seems legit to me. Still a little worried about how we know A∩D and C∪B are open in X though.

I'm not sure why. In the context of your formulation aren't $A, B, C, D$ open in $X$?