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Here is the page that is confusing me: Page 25 of General Topology (Willard)

Definition 3.5 If $X$ is a topological space, and $E\subset X$, then the closure of $E$ in $X$ is the set $$\overline{E} = \mathrm{Cl}(E) = \bigcap \{K\subset X\mid K\text{ is closed and }E\subset K\}.$$

[...]

Lemma 3.6 If $A\subset B$, then $\overline{A}\subset \overline{B}$.

[...]

Theorem 3.7 The operation $A\mapsto \overline{A}$ in a topological space has the following properties:

K-a) $E\subset \overline{E}$

K-b) $\overline{(\overline{E})}=\overline{E}$

K-c) $\overline{A\cup B} =\overline{A}\cup\overline{B}$

K-d) $\overline{\varnothing} = \varnothing$

K-e) $E$ is closed in $X$ iff $\overline{E}=E$.

Moreover, given a set $X$ and a mapping $A\mapsto \overline{A}$ of $\mathscr{P}(X)$ into $\mathscr{P}(X)$ satisfying K-a through K-d, if we define closed sets using K-e, the result is a topology on $X$ whose closure operator is just the operation $A\mapsto\overline{A}$ we began with.

Everything was fine until this page, and suddenly I'm utterly confused. Definition 3.5 defines what a closure is, and lemma 3.6 follows directly from this definition and is, it seems to me, almost self-evident... and then Theorem 3.7 happens and I've no idea what's going on.

In the second-to-last paragraph, Willard implies that, in the collection F of all sets $\bar{A}=A$, $A \subset B \implies \bar{A} \subset \bar{B}$ does not directly follow from lemma 3.6.

We proceed now to the second part of the theorem. Let $X$ be any set and $A\to \overline{A}$ a mapping of $\mathscr{P}(X)$ into $\mathscr{P}(X)$ satisfying K-a through K-d. Let $\mathscr{F}$ be the collection of all sets $A$ such that $A=\overline{A}$. The assertion is that $\mathscr{F}$ satisfies F-a through F_c of Theorem 3.4.

First note that if $A\subset B$, then by K-c, $\overline{B}=\overline{A}\cup\overline{B-A}$, so that $\overline{A}\subset \overline{B}$ (why couldn't we just refer to Lemma 3.5?)

Wouldn't it follow even more clearly so? Isn't this valid: $ \bar{A} (= A) \subset \bar{B} (= B) \implies Lemma \; 3.6$

In fact, I'm having trouble understanding the significance of this Kuratowski closure operation altogether... don't K-a to K-e all follow from *Definition 3.5?

I think I might need a more extensive article on this topic, because I don't fully understand the significance of the material on this page.

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Please be aware that google book links are highly unstable. What you are allowed to see often depends on which country you are viewing from. As such, it would be best if the post is self-contained and includes the relevant definitions and lemmas. –  Arturo Magidin Feb 24 '12 at 20:46
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About the tags: Not everything about sets is [set-theory]; questions about online resources fit into [online-resources], and not questions directing the reader to one; and reference request is a request for a reference, not asking about a specific source. –  Asaf Karagila Feb 24 '12 at 20:49
    
@ArturoMagidin Ok, thank. Wow you're fast with the LaTex. –  iDontKnowBetter Feb 24 '12 at 21:00
    
@DavidMitra so I have to pay even closer attention to initial conditions and definitions than I'm previously used to. Missing that small detail completely threw me off. –  iDontKnowBetter Feb 24 '12 at 21:38

1 Answer 1

up vote 4 down vote accepted

There are two parts to Theorem 3.7.

The first part says: if you already have a topological space $X$, and you define $\overline{A}$ as per Definition 3.5, then the operation $A\mapsto \overline{A}$ will satisfy condition K-a through K-e.

The second part of Theorem 3.7 says: suppose you have a set (not a topological space, just a set), together with a function from $\mathscr{P}(X)$ to $\mathscr{P}(X)$, denoted by $A\mapsto\overline{A}$ which just happens to satisfy conditions K-a through K-d. Then you can use this function to define a topology on $X$ by saying "$A$ is closed in $X$ if and only if $A=\overline{A}$."

Now, if you do that, you started with a function $A\mapsto\overline{A}$; now you have a topological space. Since you now have a topological space, you can use Definition 3.5 to define a closure operator on this topological space, by $$\mathrm{Cl}(A) = \bigcap \{ K\subset X\mid K\text{ is closed and }A\subset K\}.$$ If you do this, what is the relationship between $\overline{A}$ and $\mathrm{Cl}(A)$? The answer, given in the last paragraph, is that if you do this two-step process, then the function $\mathrm{Cl}(A)$ will just be exactly $\overline{A}$. That is, the function you get by starting with your original map, then constructing the topology, and then using the topology to give you a closure operator is exactly the function you started with.

The reason you cannot invoke Lemma 3.6 in that paragraph is that the paragraph is taking place in the context of the second part of the theorem. That is, you do not have a topology and you do not have the closure operator defined by the topology; you just have some function from $\mathscr{P}(X)$ to $\mathscr{P}(X)$ which happens to satisfy conditions K-a through K-d; Lemma 3.6 is about the closure operator you define once you have a topology, but since you don't have a topology here yet, you can't use it.

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Oh, I see. Thanks! –  iDontKnowBetter Feb 24 '12 at 21:22
    
I should refresh pages more often. Deleting my comments.. –  David Mitra Feb 24 '12 at 21:37
    
@DavidMitra just after I posted my reply to yours! haha, now this is like one of those Benny Hill chase scenes. –  iDontKnowBetter Feb 24 '12 at 21:41

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