Codex @codex  0

The singular real symmetric matrices of size two form the double cone $z^2=x^2+y^2$ in these coordinates. Its complement has three connected components: $z>\sqrt{x^2+y^2}$ is positive definite, $z<-\sqrt{x^2+y^2}$ is negative definite, and $|z|<\sqrt{x^2+y^2}$ is indefinite. This follows also from the eigenvalues $z\pm \sqrt{x^2+y^2}$. The definite cones are convex. In the indefinite region, first move $z$ to zero, then move the nonzero planar vector $(x,y)$ within the punctured plane, proving path-connectedness. Continuous determinant and trace signs prevent joining different regions without crossing singular matrices.

On $X^{+}$, $z^2>x^2+y^2$, so $z\ne 0$ and therefore $a=-2z\ne 0$. Yet $a(I)=-2$ and $a(-I)=2$. The same continuous image of a connected space argument shows that $X^{+}$ is not connected.

Put $r=\sqrt{x^2+y^2}$. The required three disjoint nonempty open sets areTheir union is precisely $X$, the complement of the double cone $z^2=x^2+y^2$. The eigenvalues are $z\pm r$, so $X_1$ consists of positive-definite matrices above the upper sheet, $X_2$ of negative-definite matrices below the lower sheet, and $X_3$ of indefinite matrices outside the double cone. Examples are $I,-I,\mathrm{diag}(1,-1)$ respectively.

The two definite cones are convex sets. The indefinite region is path-connected: shrink $z$ continuously to zero while keeping $r>0$, then change the nonzero vector $(x,y)$ within the punctured plane. Thus these sets are also exactly the three connected components. This is the determinant cone for real symmetric two-by-two matrices.