# Topology boundary problem

1. Oct 26, 2008

### LMKIYHAQ

1. The problem statement, all variables and given/known data

Let X be a space. A$$\subseteq$$X and U, V, W $$\in$$ topolgy(X). If W$$\subseteq$$ U$$\cup$$ V and U$$\cap$$ V$$\neq$$ emptyset,

Prove bd(W) = bd(W$$\cap$$U) $$\cup$$ bd (W$$\cap$$ V)

2. Relevant equations
bd(W) is the boundary of W...
I think I have the "$$\supseteq$$" part, but I am having trouble with the "$$\subseteq$$" part.

3. The attempt at a solution
$$\supseteq$$: Assume x $$\in$$ bd(W$$\cap$$U) $$\cup$$ bd(W $$\cap$$V). Show x$$\in$$ bd(W). Then x $$\in$$ bd(W$$\cap$$ U) or x$$\in$$ bd(W$$\cap$$ V). If x$$\in$$ bd (W$$\cap$$U) means x is in bd(W) since W$$\cap$$U$$\neq$$ emptyset and since W$$\subseteq$$ U$$\cup$$ V, some part of W$$\subseteq$$U.
If x$$\in$$ bd(W$$\cap$$ V) then x$$\in$$ bd(W) since W$$\cap$$ V$$\neq$$ emptyset and since W$$\subseteq$$U$$\cap$$ V, some part of W$$\subseteq$$ V.

Does this look ok for this part of the proof?

$$\subseteq$$: Assume x$$\in$$bd(W). Show x$$\in$$ bd(W$$\cap$$ U)$$\cup$$ (bd(W$$\cap$$ V). Since U and V are disjoint and W$$\subseteq$$U$$\cup$$, W_U$$\subseteq$$U or W_V$$\subseteq$$V. Suppose W_U$$\subseteq$$U then x$$\in$$bd(W_U)$$\subseteq$$bd(W_U$$\cap$$U).Suppose W_V$$\subseteq$$ V then x$$\in$$ bd(W_V)$$\subseteq$$bd(W_V$$\cap$$ V). Since x$$\in$$bd(W)$$\subseteq$$bd(W_U$$\cap$$U) or bd(W_V$$\cap$$ V) then x$$\in$$ bd(W$$\cap$$ U)$$\cup$$ (bd(W$$\cap$$ V).

I am not sure if I need to specify what W_U and W_V are? or if this even works for this second part of the proof?

Last edited: Oct 26, 2008
2. Oct 27, 2008

### morphism

Why is this the case?

OK, I think I see why you're having a problem:
$U \cap V \neq \emptyset$ says that U and V are not disjoint.

3. Oct 27, 2008

### LMKIYHAQ

Thanks morphism. I tried to fix it. Am I on the right track?

$$\subseteq$$: Assume x$$\in$$ bd(W). Show x$$\in$$ bd(W$$\cap$$ U)$$\cup$$ (bd(W$$\cap$$ V). Since U and V are not disjoint and W$$\subseteq$$U$$\cup$$, W_U$$\subseteq$$U or W_V$$\subseteq$$V where W_U is the set of W in U and W_V is the set of W in V. Suppose W_U$$\subseteq$$U then x$$\in$$bd(W_U)$$\subseteq$$bd(W_U$$\cap$$U).Suppose W_V$$\subseteq$$ V then x$$\in$$ bd(W_V)$$\subseteq$$bd(W_V$$\cap$$ V). Since x$$\in$$bd(W)$$\subseteq$$bd(W_U$$\cap$$U) or bd(W_V$$\cap$$ V) then x$$\in$$ bd(W$$\cap$$ U)$$\cup$$ (bd(W$$\cap$$ V).

Last edited: Oct 27, 2008
4. Oct 27, 2008

### morphism

I'm not sure I quite understand what W_U and W_V are supposed to denote.

5. Oct 27, 2008

### LMKIYHAQ

Ignore my last proof...I have been working on another approach to proving
$$\subseteq$$.

I am struggling but I think I am beginning to understand. Will you take a look?

Assume x[text]\in[/tex] bd(W). Show
x $$\in$$ bd(W$$\cap$$U) $$\cup$$
bd(W$$\cap$$V). So x$$\in$$cl(W)-int(W) and we know that W$$\subseteq$$U$$\cap$$V$$\neq$$ emptyset. Then x$$\in$$(bd(W)$$\cap$$U) $$\cup$$(bd(W)$$\cap$$V). Since bd(W)$$\subseteq$$ cl(W), x$$\in$$(cl(W)$$\cap$$U) $$\cup$$(cl(W)$$\cap$$V).

So x$$\in$$cl(W$$\cap$$U) $$\cup$$cl(W$$\cap$$V).
Since x$$\in$$bd(W), x$$\in$$bd(W$$\cap$$U) $$\cup$$bd(W$$\cap$$V).

Last edited: Oct 27, 2008