Consider the following Hasse diagram:
My book gives the following join and meet operation tables for this diagram:
$$\begin{array}{|c || c | c|} \hline Subset & x \wedge y & x \vee y \\ \hline \{a,b\} & 0 & 1 \\ \{a,c\} & c & a \\ \{a,1\} & a & 1 \\ \{a,0\} & 0 & a \\ \{b,c\} & 0 & 1 \\ \{b,a\} & 0 & 1 \\ \{b,1\} & b & 1 \\ \{b,0\} & 0 & b \\ \{c,a\} & c & a \\ \{c,1\} & c & 1 \\ \{c,b\} & b & 1 \\ \{c,0\} & 0 & c \\ \hline \end{array}$$
I don't follow, why,
$a\land c$ = $c$ and $a\lor c$ =$a$, shouldn't it be vice versa ?
$a\land b$ = $0$ ?
$a\lor b$ = $1$ ?
It would be nice if someone could help me out here.
As suggested in the comments, the table is wrong in several places. Here is a corrected table:
$$\begin{array}{|c || c | c|} \hline \text{Subset} & x \wedge y & x \vee y \\ \hline \{a,b\} & 0 & 1 \\ \{a,c\} & a & c \\ \{a,1\} & a & 1 \\ \{a,0\} & 0 & a \\ \{b,c\} & 0 & 1 \\ \{b,1\} & b & 1 \\ \{b,0\} & 0 & b \\ \{c,1\} & c & 1 \\ \{c,0\} & 0 & c \\ \hline \end{array}$$
Now why do we have $a \wedge b = 0$ and $a \vee b = 1$? In terms of the Hasse diagram, $0$ is the first element "downwards accessible" from both $a$ and $b$; likewise for $1$, but "upwards".