1. ## interior sets

Greetings,

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

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

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

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

Give number 2 a try yourself now.

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

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 $W_1=(0,1]~\&~W_2=(1,2)$.

$\text{Int} (W_1 ) = (0,1)\;,\;\text{Int} (W_2 ) = (1,2)\;\& \;\text{Int} (W_1 \cup W_2 ) = (0,2)$
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 $W_{1}$ 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 $W_{1}$ 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 $W_1\text{ and }W_2$ are anything other than any two sets.

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

6. Originally Posted by Plato
There is nothing in the statement of the question to suggest that $W_1\text{ and }W_2$ are anything other than any two sets.
So the specification in the OP that $W_{1}$ and $W_{2}$ 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 $W_{1}$ and $W_{2}$ 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 $W_{1}$ and $W_{2}$ are both arbitrary subsets of $X,$ and not necessarily members of the topology $\tau$?

9. Originally Posted by Ackbeet
So, in looking at wiki's definition of a topological space, you're saying that your $W_{1}$ and $W_{2}$ are both arbitrary subsets of $X,$ and not necessarily members of the topology $\tau$?
Well of course. Because an open set is its own interior then if $W_1~\&~W_{2}$ 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!