Let (X, O) be a topological space. Prove the following statements about the closed subsets of X:
(i) 0 and X are closed
(ii) If F1, F2 contained in X are closed, then the union of F1 and F2 is closed.
Let $\displaystyle T$ be a topology on set $\displaystyle X$.
1) Since $\displaystyle \{\}\in T$ its complement is $\displaystyle X$ thus, $\displaystyle X$ is closed.
Now since $\displaystyle X\in T$ its complement is $\displaystyle \{\}$ thus, $\displaystyle \{\}$ is closed.
2)Since $\displaystyle F_1,F_2$ are closed there exists sets $\displaystyle S_1,S_2\in T$ respectively, such as their complenets give these sets. But, because $\displaystyle T$ is topological space then the union of any sets is an element of $\displaystyle T$ (by definition).But, the complement of $\displaystyle S_1\cup S_2$ is $\displaystyle F_1\cup F_2$. Thus, $\displaystyle F_1\cup F_2$ is closed.
Just a warning I have never studied topology. Thus, what I may have said is completely wrong. But I think I understand the question.
The complement of a union is the intersection of the complements, so thisOriginally Posted by ThePerfectHacker
does not work. What does is:
$\displaystyle (F_1\cup F_2)'=F_1'\cap F_2'$,
(where $\displaystyle A'$ denotes the complement of $\displaystyle A$)
but the intersection of two open sets (in this case $\displaystyle F_1'$ and $\displaystyle F_2'$), is open so
$\displaystyle (F_1\cup F_2)'$ is open, and so $\displaystyle F_1\cup F_2$ is closed.
.
RonL
Stupid me forgot to use de Morgan's Law's for sets
CaptainBlack is right.
But you (CaptainBlack) forgot to mention one important thing when you were destroying my proof. You forgot to state the the intersection of two sets is an element of the topological space.
I don't need to as topological space (X,O) is an ordered pair where X is aOriginally Posted by ThePerfectHacker
set (space) and O is the collection of subsets of X which are open.
X being a set inherits all the properties of sets, such as intersections and
unions of subsets of X also being subsets of X.
RonL
Perhaps if you are answering a post on a topic you are not familiar withOriginally Posted by ThePerfectHacker
you should at least first look it up on MathWorld or on Wikipedia,
both of which are pretty good for this (I usualy do even if I am familiar with
the field).
When did I learn about topological spaces - as a final year undergraduate in
a course on Functional Analysis, about 30 years ago, but I don't see how
that is relevant to anything.
RonL
This is just an application of de Morgan's laws, see Alternate Definitions atOriginally Posted by TexasGirl
Topological Space on Wikipedia for an definition of a topological space in terms
of closed sets, and de Morgan's Laws on PlanetMath for the stuff needed
to justify the Wikipedia statement.
But briefy (look at the references for justification):
$\displaystyle \left( \bigcup_{a\epsilon A} F_a' \right) '=\bigcap_{a\epsilon A} F_a$,
but the LHS is the complement of an open set and so closed so the RHS is
closed (an arbitary union of open sets is open, and all the $\displaystyle F_a$ are close
so all the $\displaystyle F_a'$ are open).
RonL
I was thinking about topological spaces and wondered if this has any importance.
1)Let $\displaystyle G$ be a group.
2)Let $\displaystyle G'=\{X\subseteq G\}$
3)Define $\displaystyle S=\{\}\cup G'$
4)Thus, $\displaystyle S$ is a topology on $\displaystyle G$
This is because the union/intersection of two groups is a itself a group and also because $\displaystyle S$ satisfies the topology axioms.
A group $\displaystyle \mathcal{G}$ is an ordered pair $\displaystyle (G,\otimes)$, where $\displaystyle G$ is a set and $\displaystyle \otimes$ is aOriginally Posted by ThePerfectHacker
binary operation defined on $\displaystyle G\times G$, such that the combination satisfies
the axioms for a group.
You have only used the set $\displaystyle G$ in your construction and none of
the properties of $\displaystyle \otimes$, so what you have done is in fact true
for any set.
I presume you mean:2)Let $\displaystyle G'=\{X\subseteq G\}$
$\displaystyle G'=\{X; \mbox{such that } X \subseteq G \}$
In which case this is the power set of $\displaystyle G$, usually written:
$\displaystyle \mathcal{P} (G)$.
If my interpretation above of what you mean is correct, then this is redundant3)Define $\displaystyle S=\{\}\cup G'$
as $\displaystyle \emptyset \subset G$ and so an element of $\displaystyle \mathcal{P} (G)$ already (the null set is a subset of every set).
Yes.4)Thus, $\displaystyle S$ is a topology on $\displaystyle G$
This is because the union/intersection of two groups is a itself a group and also because $\displaystyle S$ satisfies the topology axioms.
RonL
CaptainBlack what you said is true. I made a mistake in how I posed my question, I wanted to say a subgroup not a subset.
The revised question:
1)Let $\displaystyle G$ be a group.
2)Let $\displaystyle P(G)=\{X|X\leq G\}$.
3)Let $\displaystyle S=\{\} \cup P(G)$.
4)Then $\displaystyle S$ is a topology on $\displaystyle G$.
The reason is because the union/intersection of two subgroups is a subgroup.