1. ## interior sets

Greetings,

I have two questions that I hope someone can assist me with. We let $\displaystyle W_{1}$ and $\displaystyle W_{2}$ be arbitrary subsets in the topological space $\displaystyle M$. I want to show that:

1) $\displaystyle int \left(W_{1} \cap W_{2} \right)=int \left(W_{2}\right) \cap int\left(W_{2}\right)$

2) $\displaystyle int \left(W_{1} \cup W_{2} \right)\subseteq int \left(W_{2}\right) \cup int\left(W_{2}\right)$

I would greatly appreciate it if someone could help me through the problem so that I understand it and not just give me the answer.

Thanks.

2. (1) Let x be a point in the interior of $\displaystyle W_1 \cap W_2$. Then x is the center of an open ball $\displaystyle B_1$ fully contained in $\displaystyle W_1$ and a ball $\displaystyle B_2$ fully contained in $\displaystyle W_2$. Let $\displaystyle B$ be the intersection of these two balls. $\displaystyle B$ is in the interior of $\displaystyle W_1$ and the interior of $\displaystyle W_2$.

Give number 2 a try yourself now.

(By the way, I'm pretty sure you have 2 typos in the question - 2 of the $\displaystyle W_2$ should be $\displaystyle W_1$)

3. Remark: This is a topological space not necessarily metric space. It would be better to say open set in stead open ball.

There is more wrong with #2 that just a typo.
In space of real numbers with the usual topology let $\displaystyle W_1=(0,1]~\&~W_2=(1,2)$.

$\displaystyle \text{Int} (W_1 ) = (0,1)\;,\;\text{Int} (W_2 ) = (1,2)\;\& \;\text{Int} (W_1 \cup W_2 ) = (0,2)$
Do you see something wrong now?

4. Reply to Plato at #3:

I never took topology, so forgive me if I'm wrong. But isn't your $\displaystyle W_{1}$ not in the usual topology on the reals? It's not an open set in the usual topology, right? So the theorem wouldn't apply?

5. Originally Posted by Ackbeet
I never took topology, so forgive me if I'm wrong. But isn't your $\displaystyle W_{1}$ not in the usual topology on the reals?
This is a standard exercise in basic topology.

There is nothing in the statement of the question to suggest that $\displaystyle W_1\text{ and }W_2$ are anything other than any two sets.

The statement should be $\displaystyle \text{Int}(W_1)\cup \text{Int}(W_2)\subseteq \text{Int}(W_1\cup W_2)$.

6. Originally Posted by Plato
There is nothing in the statement of the question to suggest that $\displaystyle W_1\text{ and }W_2$ are anything other than any two sets.
So the specification in the OP that $\displaystyle W_{1}$ and $\displaystyle W_{2}$ are arbitrary subsets in the topological space M does not imply that they are open? I thought every subset in a topological space had to be open - indeed, was defined to be open.

7. Originally Posted by Ackbeet
So the specification in the OP that $\displaystyle W_{1}$ and $\displaystyle W_{2}$ are arbitrary subsets in the topological space M does not imply that they are open? I thought every subset in a topological space had to be open - indeed, was defined to be open.
Do not confuse these two:
1. A subset of a topological space.
2. A set in the topology of the space.

The first means that an arbitrary set in the space and the second means that the set is in fact open (the set in the topology are open).

8. So, in looking at wiki's definition of a topological space, you're saying that your $\displaystyle W_{1}$ and $\displaystyle W_{2}$ are both arbitrary subsets of $\displaystyle X,$ and not necessarily members of the topology $\displaystyle \tau$?

9. Originally Posted by Ackbeet
So, in looking at wiki's definition of a topological space, you're saying that your $\displaystyle W_{1}$ and $\displaystyle W_{2}$ are both arbitrary subsets of $\displaystyle X,$ and not necessarily members of the topology $\displaystyle \tau$?
Well of course. Because an open set is its own interior then if $\displaystyle W_1~\&~W_{2}$ were in the topology they would be open so there would be nothing to prove.

10. Got it. Thanks for letting me play the incompetent critic. I learned something!