How Can You Prove the Boundary of a Set in Topology?

[LMKIYHAQ]

Homework Statement

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

Prove bd(W) = bd(W\capU) \cup bd (W\cap V)

Homework Equations

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

The Attempt at a Solution

\supseteq: Assume x \in bd(W\capU) \cup bd(W \capV). Show x\in bd(W). Then x \in bd(W\cap U) or x\in bd(W\cap V). If x\in bd (W\capU) means x is in bd(W) since W\capU\neq emptyset and since W\subseteq U\cup V, some part of W\subseteqU.
If x\in bd(W\cap V) then x\in bd(W) since W\cap V\neq emptyset and since W\subseteqU\cap V, some part of W\subseteq V.

Does this look ok for this part of the proof?

\subseteq: Assume x\inbd(W). Show x\in bd(W\cap U)\cup (bd(W\cap V). Since U and V are disjoint and W\subseteqU\cup, W_U\subseteqU or W_V\subseteqV. Suppose W_U\subseteqU then x\inbd(W_U)\subseteqbd(W_U\capU).Suppose W_V\subseteq V then x\in bd(W_V)\subseteqbd(W_V\cap V). Since x\inbd(W)\subseteqbd(W_U\capU) 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?

 

[morphism]

The Attempt at a Solution

\supseteq: ... since W\capU\neq emptyset
... since W\cap V\neq emptyset ...

Why is this the case?

OK, I think I see why you're having a problem:

Since U and V are disjoint

U \cap V \neq \emptyset says that U and V are not disjoint.

 

[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\subseteqU\cup, W_U\subseteqU or W_V\subseteqV where W_U is the set of W in U and W_V is the set of W in V. Suppose W_U\subseteqU then x\inbd(W_U)\subseteqbd(W_U\capU).Suppose W_V\subseteq V then x\in bd(W_V)\subseteqbd(W_V\cap V). Since x\inbd(W)\subseteqbd(W_U\capU) or bd(W_V\cap V) then x\in bd(W\cap U)\cup (bd(W\cap V).

 

[morphism]

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

 

[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\capU) \cup
bd(W\capV). So x\incl(W)-int(W) and we know that W\subseteqU\capV\neq emptyset. Then x\in(bd(W)\capU) \cup(bd(W)\capV). Since bd(W)\subseteq cl(W), x\in(cl(W)\capU) \cup(cl(W)\capV).

So x\incl(W\capU) \cupcl(W\capV).
Since x\inbd(W), x\inbd(W\capU) \cupbd(W\capV).