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====== Topology, 2ed ====== | ====== 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 $\subseteq$, | ||
+ | |||
+ | 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 // | ||
+ | |||
+ | Composites, preimages, injectivity, | ||
+ | |||
+ | ++++ 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), | ||
+ | |||
+ | (b) | ||
+ | ++++ | ||
+ | |||
+ | ++++ Exercise 2.2 | | ||
+ | > Let $f:A\to B$, and let $A_i\subseteq A$ and $B_i\subseteq B$ for $i\in\{0, | ||
+ | > (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**: | ||
+ | |||
+ | 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)< | ||
+ | </ | ||
+ | |||
+ | <WRAP center round box 80%> | ||
+ | **Definition**: | ||
+ | |||
+ | 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 center round box 80%> | ||
+ | **Definition**: | ||
+ | |||
+ | 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 // | ||
+ | |||
+ | 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 // | ||
+ | </ | ||
+ | |||
+ | 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, | ||
+ | |||
+ | ++++ Exercise 3.13 | | ||
+ | > Suppose that an ordered set $A$ has the least upper bound property. Prove that it has the greatest lower bound property. | ||
+ | |||
+ | ++++ | ||
+ | |||
+ | |||
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