Set operations কী?

Example 1.17 (Getting used to the power set)

For $E = \{a, b\}$:

four elements — and note the type discipline: $a \in E$ but $\{a\} \in \mathcal P(E)$; the statements $a \in \mathcal P(E)$ and $\{a\} \subseteq \mathcal P(E)$ are both false as written (the second would require $a$ to be a subset of $E$). Iterating from nothing: $\mathcal P(\emptyset) = \{\emptyset\}$ has one element, $\mathcal P(\mathcal P(\emptyset)) = \{\emptyset, \{\emptyset\}\}$ has two, the next has four — sets of sets are ordinary sets, and Chapter 2 will confirm the doubling pattern: $\abs{\mathcal P(E)} = 2^{\abs E}$. Keeping the levels ($x$, $\{x\}$, $\{\{x\}\}$) straight is half the battle in exercises like Exercises 1.11 and 1.12.

Example 1.27 (Point (2) is sharp)

In Proposition 1.26 (2), one cannot upgrade the conclusions: $g \circ f$ bijective does not force $f$ surjective or $g$ injective. Take $E = G = \{1\}$, $F = \{1, 2\}$, with $f(1) = 1$ and $g(1) = g(2) = 1$: then $g \circ f = \mathrm{id}_E$ is bijective, yet $f$ misses the element $2$ and $g$ collapses both elements. The moral is a precise bookkeeping rule: composition information flows to the inner map for injectivity and to the outer map for surjectivity, never the other way. (Exercise 1.9 builds the same phenomenon with infinite sets, where it is the engine behind one-sided inverses.)

Example 1.32

On $\Z$, congruence modulo $n$ ($x \equiv y \pmod n$ when $n$ divides $x - y$) is an equivalence relation; its classes are the $n$ sets of integers with a given remainder upon division by $n$. This example becomes the ring $\Z/n\Z$ in Chapter 7.