====== Topology, 2ed ======

===== Notes =====
==== Set Theory and Logic ====
This chapter is a review of basic mathematical concepts, beginning with set theory. One key point to keep in mind is that Munkres will occasionally disambiguate the notation $(a,b)$ as an ordered pair versus as an interval by using $a\times b$ for the former. The text also prefers $\subset$ over $\subseteq$, emphasizing proper subsets with $\subsetneq$. I will not follow that convention in these notes. 

Given a function $f:A\to B$, the text defines:

  * its //domain// as $A$
  * its //range// as $B$
  * its //image set// as $f(A)$
  * the //restriction// of $f$ to $A_0\subseteq A$, denoted $f|A_0$, as $\{(a, f(a)): a\in A_0\}$

Composites, preimages, injectivity, surjectivity, and bijectivity are all defined as usual.

++++ Exercise 2.1 |
> Let $f:A\to B, A_0\subseteq A, B_0\subseteq B$.
> (a) Show that $A_0\subseteq f^{-1}(f(A_0))$ and that equality holds if $f$ is injective.
> (b) Show that $f(f^{-1}(B_0))\subseteq B_0$ and that equality holds if $f$ is surjective. 

(a) Let $a\in A_0$. $a=f^{-1}(f(a))\in f^{-1}(f(A_0))\implies A_0\subseteq f^{-1}(f(A_0))$. Now suppose $f$ is injective, and let $a\in f^{-1}(f(A_0))$. $\exists b\in f(A_0)$ s.t. $a=f^{-1}(b)$. Since $b\in f(A_0),\exists a^\prime\in A_0$ s.t. $f(a^\prime)=b$. Since $f$ is injective, we must have $a=a^\prime$; that is, $a\in A_0$. Hence $f^{-1}(f(A_0))=A_0$.

(b) 
++++

++++ Exercise 2.2 |
> Let $f:A\to B$, and let $A_i\subseteq A$ and $B_i\subseteq B$ for $i\in\{0,1\}$. Show that:
> (a) $B_0\subseteq B_1 \implies f^{-1}(B_0)\subseteq f^{-1}(B_1)$
> (b) $f^{-1}(B_0\cup B_1) = f^{-1}(B_0)\cup f^{-1}(B_1)$
> (c) $f^{-1}(B_0\cap B_1) = f^{-1}(B_0)\cap f^{-1}(B_1)$
> (d) $f^{-1}(B_0 - B_1) = f^{-1}(B_0) - f^{-1}(B_1)$
> (e) $A_0\subseteq A_1\implies f(A_0)\subseteq f(A_1)$
> (f) $f(A_0\cup A_1)=f(A_0)\cup f(A_1)$
> (g) $f(A_0\cap A_1)\subseteq f(A_0)\cap f(A_1)$; equality holds if $f$ is injective
> (h) $f(A_0 - A_1)\subseteq f(A_0) - f(A_1)$; equality holds if $f$ is injective

++++

<WRAP center round box 80%>
**Definition**: order type

Suppose that $A$ and $B$ are two sets with respective order relations $<_A$ and $<_B$. We say that $A$ and $B$ have the same //order type// if there exists a bijection $f:A\to B$ s.t. $a_1<_A a_2 \implies f(a_1)<_B f(a_2)$.
</WRAP>

<WRAP center round box 80%>
**Definition**: largest element, smallest element

Suppose that $A$ is a set ordered by the relation $<$. Let $A_0\subseteq A$. We say that $b$ is the //largest element// of $A_0$ if $b\in A_0$ and $x\leq b~\forall x\in A_0$. Similarly, we say that $a$ is the //smallest element// of $A_0$ if $a\in A_0$ and $a\leq x~\forall x\in A_0$. 
</WRAP>

<WRAP center round box 80%>
**Definition**: bounded above, upper bound, supremum, bounded below, lower bound, infimum

We say that the subset $A_0$ is //bounded above// if there is an element $b\in A$ s.t. $x\leq b~\forall x\in A_0$; $b$ is called an //upper bound// for $A_0$. If the set of all upper bounds for $A_0$ has a smallest element, that element is called the //supremum//, denoted $\sup A_0$. 

Similarly, $A_0$ is //bounded below// if there is an element $a\in A$ s.t. $a\leq x~\forall x\in A_0$; $a$ is a //lower bound// for $A_0$. If the set of all lower bounds of $A_0$ has a largest element, that element is called the //infimum//, denoted $\inf A_0$. 
</WRAP>

An ordered set $A$ has the //least upper bound property// if each nonempty subset $A_0$ of $A$ that is bounded above has a least upper bound. Analogously, it has the //greatest lower bound property// if each nonempty $A_0\subseteq A$ which is bounded below has a greatest lower bound. $A$ has one of these properties iff it has the other (see Exercise 3.13). 

++++ Exercise 3.13 |
> Suppose that an ordered set $A$ has the least upper bound property. Prove that it has the greatest lower bound property.

++++



{{tag>math topology algebra}}