# Thread: Proof: Interior points and closed sets.

1. ## Proof: Interior points and closed sets.

I need help with another complex problem in a general topological space:

Show that a set S is open if and only if each point in S is an interior point.

2. Originally Posted by Porter1
I need help with another complex problem in a general topological space:
Show that a set S is open if and only if each point in S is an interior point.
If each point of a set $\mathcal{O}$ is an interior point then for all $x \in \mathcal{O}$ there is an open set $Q_x\subseteq \mathcal{O}$.
Prove that $\bigcup\limits_{x \in \mathcal{O}} {Q_x } = \mathcal{O}$

3. Thats the thing, I dont know how to prove it.

4. Originally Posted by Porter1
Thats the thing, I dont know how to prove it.
Please believe me, I mean you no disrespect.
Have you ever considered that you may not have the necessary preparation to tackle this material?
Is a retired chair of a mathematical sciences department, I have had this conversation with a countless number of students.

5. Originally Posted by Porter1
I need help with another complex problem in a general topological space:

Show that a set S is open if and only if each point in S is an interior point.
In the other direction to Plato's without all the details. Let $O$ be an open set in a topological space $X$. Suppose that $x\in O$ was not an interior point, then every neighborhood $N$ of $x$ intersects $X-O$. Therefore, we see that $x$ is a limit point of $O'$. But, we have that $x$ is a limit point of $O'$ that is not in $O'$, so $O'$ is not closed and thus $O$ is not open. This, of course, is a contradiction.

This is an indirect way. Of course it assumes you've defined closed set and limit point. If not, you should be able to see a more direct way.

6. Originally Posted by Porter1
I need help with another complex problem in a general topological space:

Show that a set S is open if and only if each point in S is an interior point.

By writing down two definitions the above statement follows quite easily:

Def 1 x is interior point of S <=> there exists ε>0 and $B(x,\epsilon)\subset S$

Def 2 S is open <=> for all ,x: xεS => there exists ε>0 and $B(x,\epsilon)\subset S$.

And using the two definitions we have:

S is open <=> (for all ,x : xεS => there exists ε>0 and $B(x,\epsilon)\subset S$) <=> (for all ,x : xεS => x is interior of S) <=> (each point of S is interior)

7. Originally Posted by xalk

By writing down two definitions the above statement follows quite easily:

Def 1 x is interior point of S <=> there exists ε>0 and $B(x,\epsilon)\subset S$

Def 2 S is open <=> for all ,x: xεS => there exists ε>0 and $B(x,\epsilon)\subset S$.

And using the two definitions we have:

S is open <=> (for all ,x : xεS => there exists ε>0 and $B(x,\epsilon)\subset S$) <=> (for all ,x : xεS => x is interior of S) <=> (each point of S is interior)
This is not a metric space necessarily. So, while there are still neighborhoods the idea of an open ball is not present.

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