Thread: Topology Boundary of a set is closed? A set subset of it's interior implies open set?

1. Topology Boundary of a set is closed? A set subset of it's interior implies open set?

A is a nonempty set. Show boundary of A is closed. Obviously dealing in the real number space.

I've seen a couple of proofs for this, however they involve 'neighborhoods' and/or metric spaces and we haven't covered those. Here are some following terms we've define: segment, limit point, boundary point, interior point, boundary, derived set( the set of all limit points), isolated point, density, open, closed, interior, exterior, connected, disconnected.

I've started with this.. not much of a proof, just a thought process.

The boundary of A is the set of points that are both limit points of A and AC . Where Ac is A complement.
(i.e. boundary of A is the derived set of A intersect the derived set of Ac )

Note: boundary of A is closed if and only if every limit point of boundary of A is in boundary of A. (?or in boundary of the derived set of A is open?)

- If you could lead me in the right direction or show me how to go out the proof, that would be rad to say the least. Thanks.

Another one if you care to:

A is a subset of the interior of A. Show A is open.

All I know is a set is open if and only if for all x in A there exists a segment S containing x such that S is a subset of A.

Thank you

2. Re: Topology Boundary of a set is closed? A set subset of it's interior implies open

Originally Posted by MP91
A is a nonempty set. Show boundary of A is closed. Obviously dealing in the real number space.
The easy way to do this is to work with the basic definition of boundary point.
The statement that $x$ is a boundary point of $A$ means that if $O$ is an open interval in $\mathbb{R}$ and $x\in O$ then $O$ contains a point of $A$ and a point not in $A$.

Lets use $\beta(A)$ to be boundary of $A$.
We show that $\beta(A)^c$ is open.
Suppose that $t\notin\beta(A)$. Then by negating the definition we get, there is a open set $Q$ such that $t\in Q$ and $Q\subset A\text{ or }Q\subset A^c$. So $Q\cap\beta(A)=\emptyset.$
How does that prove that $\beta(A)$ is closed?

3. Re: Topology Boundary of a set is closed? A set subset of it's interior implies open

You could continue your approach to the proof if you observe that, for any set D, a limit point of (all limit points of D) is itself a limit point of D.
Then a limit point of limit points of A is a limit point of A, and a limit point of limit points of A-complement is a limit point of A-complement.
Do you see how that would get you your result?

So why, for any set D, a limit point of (limit point of D) is itself a limit point of D? Intuititvely it's true, but why? Can you prove it? Maybe this has already been proven in your class?

Here's one direct proof:
Proposition: Let L be the limit points of D, and x be a limit point of L. Then x is a limit point of D.

Proof:
Let r be any positive real number.
Since x is a limit point of L, B(x, r) intersect (L - {x}) is not empty.
Let x1 in B(x, r) intersect (L - {x}).

Note: x1 not equal to x, so dist(x1, x) >0.
Note: x1 in B(x, r), so dist(x1, boundary of B(x, r)) = r - dist(x1, x) > 0. (These are open balls).

Let s = MIN { dist(x1, boundary of B(x, r)), dist(x1, x) } = MIN { r - dist(x1, x), dist(x1, x) } > 0.

Note that x not in B(x1, s), since s <= dist(x1, x).
Note that, since s <= dist(x1, boundary of B(x, r)), must have that B(x1, s) is a subset of B(x, r) (use triangle inequality).

Now: x1 is in L, so it's a limit point of D, and hence B(x1, s) intersect (D - {x1}) is not empty.
Thus there exists y in B(x1, s) intersect (D - {x1}).

Since y in B(x1, s), have that y is not x.
Since y in (D - {x1}), have that y is in D.
Thus y is in (D-{x}).

Also, y in B(x1, s) which is a subset of B(x, r), so y is in B(x, r).

Thus y is in (D-{x}) intersect B(x, r).
Thus (D-{x}) intersect B(x, r) is not empty.
Since that holds for any positive real number r, it follows that x is a limit point of D.

4. Re: Topology Boundary of a set is closed? A set subset of it's interior implies open

Originally Posted by Plato
The statement that $x$ is a boundary point of $A$ means that if $O$ is an open interval in $\mathbb{R}$ and $x\in O$ then $O$ contains a point of $A$ and a point not in $A$.
My definition for boundary point is 'x' is a boundary point of A (subset of R) if and only if x is a limit point of A and x is a limit point of Ac. I'm not sure.. perhaps this is equivalent to what you are saying, but we are not giving a definition for an 'open interval', just one for an interval and an open set.

Also.. this Topology subject has been giving me much trouble. Despite having all the definitions, I really haven't been able to get a good grasp on it. Any suggestions? Perhaps helpful resources you've used in the past?

5. Re: Topology Boundary of a set is closed? A set subset of it's interior implies open

Originally Posted by MP91
My definition for boundary point is 'x' is a boundary point of A (subset of R) if and only if x is a limit point of A and x is a limit point of Ac. I'm not sure.. perhaps this is equivalent to what you are saying, but we are not giving a definition for an 'open interval', just one for an interval and an open set.
You have been given an incorrect definition of boundary point.
Here is a simple counter-example.
Let $A=\{0\}\cup [1,2]$. Now $0$ is a boundary point of $A$ but it is not a limit point of $A$

6. Re: Topology Boundary of a set is closed? A set subset of it's interior implies open

Originally Posted by johnsomeone
Proposition: Let L be the limit points of D, and x be a limit point of L. Then x is a limit point of D.
Because real numbers are a metric space and therefore a $T_2$ space it makes no difference here. But that statement is not valid in all topological spaces.

7. Re: Topology Boundary of a set is closed? A set subset of it's interior implies open

Originally Posted by MP91
Another one if you care to:
A is a subset of the interior of A. Show A is open.
All I know is a set is open if and only if for all x in A there exists a segment S containing x such that S is a subset of A.
In the real numbers a basic ball is an open segment:
$Q_x=(x-\delta,x+\delta)$ for some $\delta>0$.

If $x\in \mathcal{I}(A)$, the interior of $A$, then some $O_x\subset A$. That is the definition of interior point.

So $A=\mathcal{I}(A)$ is open.