Let
\begin{align} A &= \{(x,y) \in \mathbb R^2 : - 1 \leq x \leq 1 \land -1 \leq y \leq 1\}, \\ B &= \{(x,y) \in \mathbb R^2 : x^2 + y^2 = 1\}, \text{ and} \\ C &= \{(x,y) \in A : |x| = 1 \lor |y| = 1\}. \end{align}
How can i find the Hausdorff distance $h(B,C)$? I'm completely confused with the definition that $h(B,C) = \max\{ \sup_{x\in B} d(x,C), \sup_{x\in C} d(x,B) \}$. An explanation would be very helpful.
For every point on the circle $B$, we can find a point on the square $C$ that is at most $1-\frac12\sqrt 2$ away, the worst case happening for $(\frac12\sqrt2,\frac12\sqrt 2)\in B$ with nearest point $(\frac12\sqrt 2,1)\in C$ (plus symmetric similar cases). This means that if we "blur" $C$, or draw it with a thick pen of radius $1-\frac12\sqrt 2$, the blurred/thickend image of $C$ will cover all of $B$.
On the other hand, for each point in $C$, we can find a point in $B$ that is at most $\sqrt 2-1$ away, this time the worst case being $(1,1)\in C$ with nearest point $(\frac12\sqrt2,\frac12\sqrt 2)\in B$. (It is of no relevance and in general not the case that the same $C$-points occur in booth cases.) This means that if we "blur" $B$, or draw it with a thick pen of radius $\sqrt 2-1$, the blurred/thickened image of $B$ will cover all of $C$.
Now $h(B,C)$ is simply the larger of the two numbers $h=1-\frac12\sqrt 2$ and $j=\sqrt 2-1$.$$\boxed{h(B,C)=\max\{h,j\}}$$