TITLE: If $A\cup B = U$ and $A\cap B = \emptyset$ then $B = A^c$ -- contradiction?
QUESTION [0 upvotes]: I somehow got a weird result when proving it.
Suppose $A \cup B = U$. Then by de Morgan's Law,
$A^c \cap B^c = \emptyset$.
Taking the intersection of both sides with $A$,
$A\cap A^c \cap B^c = A \cap \emptyset = \emptyset$.
$U \cap B^c = \emptyset$
$B^c = \emptyset$?
$U$ is the universal set
Where did I go wrong here?

REPLY [0 votes]: You need to take the union of $A^C\cap B^C=\emptyset$ with $A$ giving $A\cup(A^C\cap B^C)=A\cup \emptyset$. Thus $A\cup B^C=A$ and hence $B^C\subseteq A$. Similarly, $A\subseteq B^C$.


TITLE: Solve $30u+26 \equiv 3\, \mod 7$
QUESTION [2 upvotes]: Question

Solve $30u+26 \equiv 3\,\mod 7$

I know the solution is here,but i want to know what is flaw in my approach.
My Approach

$30u+26 \equiv 3\, \mod 7$
$30u \equiv -23\,\mod 7$
$u \equiv -23\times30^{-1}\, \mod 7 \tag{1}$

let's us find $30^{-1}\, \mod 7$

$30=7\times4+2$
$7=2\times3+1$
$1=7-2\times3$
$1=7-3\times(30-7\times4)$
$1=13\times7-3\times30-$

$30^{-1}\,\mod 7=3$
putting the value of $30^{-1}\,\mod 7$ in $(1)$,

$u \equiv -23\times3\, \mod 7$
$u \equiv -69\,\mod 7=1$
but answer is $6$.  Where am I wrong?

REPLY [0 votes]: When you want to solve a linear modulo n problem, say to yourself two things:
$\quad$ I can't stand seeing negative signs.
$\quad$ I can't stand seeing numbers $\ge n$.
A chain of equivalent equations:
$30u+26 \equiv 3\, \mod 7$
$2u+5 \equiv 3\, \mod 7$
$2u \equiv 3 -5\, \mod 7$
$2u \equiv -2\, \mod 7$
$2u \equiv 5\, \mod 7$
$u \equiv (2^{-1}) (5)\, \mod 7$
$u \equiv (4) (5)\, \mod 7$
$u \equiv 20\, \mod 7$
$u \equiv 6\, \mod 7$


TITLE: (Royden Real Analysis) How is this set function $m^{**}$ related to outer measure $m^*$?
QUESTION [3 upvotes]: The following question is taken from page $43$ in the $4$th edition 'Real Analysis' by Royden, questoin $22$.
For any set $A$, define $m^{**}(A) \in [0, \infty]$ by
$m^{**}(A) = \inf \{m^*(O) : O\supset A, O \text{ is open}.\}$
How is this set function $m^{**}$ related to outer measure $m^*$?
I manage to prove that for any set $A \subseteq \mathbb{R}$, $m^{**}(A) = m^*(A)$ when both $m^{**}(A)$ and $m^{*}(A)$ are finite, that is, by considering $m^{**}(A) \leq m^{*}(A) + \varepsilon$ and $m^{*}(A) \leq m^{**}(A) + \varepsilon$.
However, I have problem proving they are equal when at least one of them is infinite.
If $m^*(A) = \infty$, then there exists a collection of disjoint measurable sets $\{ A_k \}_{k \geq 1}$ such that $A = \cup_{k \geq 1}{A_k}$ and $m^*(A_k) < \infty$ for all $k$.
From here, i couldn't proceed to show that $m^*(A) = m^{**}(A)$.
Any hint would be appreciated.

REPLY [1 votes]: If $A \subset B$ then $m^*(A) \leqslant m^*(B)$ holds in general. If $m^*(A) = +\infty$,  $O$ is open and $A \subset O$, then $m^*(O) \geqslant m^*(A) = + \infty.$
Thus, $m^*(A) = +\infty \implies m^{**}(A) = +\infty.$
The reverse implication follows in that if $m^{**}(A) = +\infty$ then for any open set $O$ with $A \subset O$, we have $m^*(O) = + \infty$. In particular this holds true for any countable union of open intervals covering $A$, whence $m^*(A) = +\infty$.