Topology Connectedness Question

**themediocrehacker** · February 21st 2011, 11:35 AM

Is it true, given A,B as two connected subsets of a topological space X s.t.
A intersection closure(B) is non-empty, that:
If C is an open & closed subset of AUB, then CU((AUB)\C)=AUB
I am not sure as both C & ((AUB)\C) are open, so is boundary(C) in either of these open sets?

If not, can you please suggest any alternative routes to tackle this problem:

Given A,B as two connected subsets of a topological space X s.t.
A intersection closure(B) is non-empty, prove that AUB is connected.

(I have tried to show that the empty set and AUB are the only two open & closed subsets of AUB by contradiction (supposing there exists C with this property))

Any help would be much appreciated,
TMH

**Plato** · February 21st 2011, 12:01 PM

Originally Posted by themediocrehacker

Given A,B as two connected subsets of a topological space X s.t. A intersection closure(B) is non-empty, prove that AUB is connected.

In the future, always start the posting with the question.
I wasted time before I came to the actual question.

Suppose that $A\cup B$ is not connected.
Then there are two nonempty separated sets $U~\&~V$ such that $A\cup B\subseteq U\cup V$ .
WLOG we can say that being connected $A\subseteq V ~\&~ B\subseteq U$ . Connected sets cannot be separated.
But not point of $U$ is a point or a limit point of $V$ .
There is a contradiction there. What is it?

**themediocrehacker** · February 21st 2011, 12:13 PM

Originally Posted by Plato

In the future, always start the posting with the question.
I wasted time before I came to the actual question.

Suppose that $A\cup B$ is not connected.
Then there are two nonempty separated sets $U~\&~V$ such that $A\cup B\subseteq U\cup V$ .
WLOG we can say that being connected $A\subseteq V ~\&~ B\subseteq U$ . Connected sets cannot be separated.
But not point of $U$ is a point or a limit point of $V$ .
There is a contradiction there. What is it?

Sorry to hear that you feel you wasted your time but the reason why I put my approach at the top was to help you pick up along the direction I was taking with this question because we haven't studied separted sets yet. If you have any ideas how to solve this using the definitions for (path) connectedness and closure I would really appreciate it

**Plato** · February 21st 2011, 12:19 PM

Originally Posted by themediocrehacker

Sorry to hear that you feel you wasted your time but the reason why I put my approach at the top was to help you pick up along the direction I was taking with this question because we haven't studied separted sets yet. If you have any ideas how to solve this using the definitions for (path) connectedness and closure I would really appreciate it

I thought that you were studying the concept of connectedness in a topological space?
That is defined by separated sets.

What definition have you been given?

**themediocrehacker** · February 21st 2011, 01:42 PM

Definiton:
Let X be a topological space.
X is connected if it satisfies one of the 3 properties below:

1) If {U,V} is an open cover of X such that U doesn't intersect V then either U is empty or V is empty.

2) The only subsets of X which are both open and closed in X are empty set and X.

3) Any continuous map from X to {0,1} (with the discrete topology) is constant

**Plato** · February 21st 2011, 03:03 PM

Well that approach is used by some authors. It is not as instructive as is the separated sets approach. But they are equivalent.

Stated properly your text’s approach ought to say:
A set $X$ is not connected if and only if there exists two disjoint open sets $U~\&~V$ each having non-empty intersection with $X$ and $X\subseteq U\cup V$

In you problem if that is true for $A\cup B$ then because each of $A~\&~B$ is connected then we can say that $A\subseteq U~\&~ B\subseteq V$ or visa versa. WHY?

But the given tells us that $A\cap \overline{B}\ne \emptyset$ .

So you have a contradiction. HOW?
Hence $A\cup B$ is connected.

**themediocrehacker** · February 22nd 2011, 02:22 AM

Originally Posted by Plato

Well that approach is used by some authors. It is not as instructive as is the separated sets approach. But they are equivalent.

Stated properly your text’s approach ought to say:
A set $X$ is not connected if and only if there exists two disjoint open sets $U~\&~V$ each having non-empty intersection with $X$ and $X\subseteq U\cup V$

In you problem if that is true for $A\cup B$ then because each of $A~\&~B$ is connected then we can say that $A\subseteq U~\&~ B\subseteq V$ or visa versa. WHY?

I'm unsure of this, I get because U union V is an open cover that is disjoint for both A and B then either U or V is empty so that either A,B is a subset of U or A,B is a subset of V

But the given tells us that $A\cap \overline{B}\ne \emptyset$ .

So you have a contradiction. HOW?
Hence $A\cup B$ is connected.

TMH

**Plato** · February 22nd 2011, 08:05 AM

The whole idea is this: suppose $U~\&~V$ are disjoint open set such that each has a non-empty intersection with $A\cup B$ and $(A\cup B)\subseteq(U\cup V)$ .

We know that $A$ is connected.
If $A\cap U\ne\emptyset~\&~ A\cap V\ne\emptyset$ that would be a disconnection of $A$ .
So $A\subset U\text{ or } A\subset V$ .

**themediocrehacker** · February 22nd 2011, 08:30 AM

Ok, I thought we were using the open cover defn there, but I can see how your definition works too.
But doesn't this imply A intersection B is empty, when we need A intersection closure(B) to be empty, for the req'd contradiction?

Thanks again for your help - I think the underlying difficulty with this topic may be that it is a extremely new concept for me.

TMH

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