Cardinality of $\{ (x, y) \in \mathbb{R}^2 \mid \left| x \right| + \left| y \right| = 1 \}$ and $\{ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1 \}$ Do $\{ (x, y) \in \mathbb{R}^2 \mid \left| x \right| + \left| y \right| = 1 \}$ and $\{ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1 \}$ have the same cardinality?
One can draw a square in the two dimensional Cartesian plane, with vertices at $(1, 0), (0, 1), (-1, 0), (0, -1)$ to represent the first set. An origin centered unit circle circumscribing the square would represent the second set. The line joining the origin to each point on the circle intersects the square at a corresponding unique point. As such, it seems from a geometrical point of view that the two sets indeed have the same cardinality.
Is there a way to rigorously show/refute that the two sets have the same number of elements (cardinality)?
 A: Define 
$$ \def\norm#1#2{\left\|#1\right\|_{#2}}\norm x1 := \def\abs#1{\left|#1\right|}\abs{x_1} + \abs{x_2} $$
and $$ \norm x2 := \left(x_1^2 + x_2^2\right)^{1/2} $$
Then your two sets are $S_1 := \{x \in \def\R{\mathbf R}\R^2 \mid \norm x1 = 1\}$ and $S_2 := \{x \in \R^2 \mid \norm x2 = 1\}$. Define maps $f \colon S_1 \to S_2$, $g \colon S_2 \to S_1$ by 
$$ f(x) := \frac{x}{\norm x2}, \qquad g(x) := \frac x{\norm x1} $$
Then $f\circ g = \mathrm{id}_{S_2}$ and $g \circ f = \mathrm{id}_{S_1}$, so $S_1$ and $S_2$ have the same cardinality.
A: HINT:
Note that both sets are subsets of $\Bbb R^2$, so their cardinality is at most that of $\Bbb R^2$, which in turn is the same as that of $\Bbb R$, and the same as $(0,1)$.
So it would suffice to find an injection from $(0,1)$ into each of these sets in order to establish they have the same size.

You can also do it in a total overkill way.
Both sets are Borel sets in a Polish space, and therefore it suffices to establish they are uncountable in order to prove they have the same cardinality as $\Bbb R$. 
But since both sets are connected subspaces of a perfectly normal, and a connect normal space with more than one point is uncountable, both sets are uncountable.
