Strict feasibility, boundary feasibility, and a weakly infeasible SDP

The slice $X_{11}=1$, $X_{22}=0$ is feasible at $X=\mathrm{diag}(1,0)$ but has no positive-definite point: a symmetric matrix in the slice is $X(a)=\left(\begin{array}{cc}1& a\\ a& 0\end{array}\right)$. If $a\mathrm{\ne }0$, the vector $(-a,1)$ gives $(-a,1)X(a)(-a,1{)}^T=-a^2<0$, so $X(a)$ is not positive semidefinite; if $a=0$, it is singular. Its Phase-I margin with $e=I$ is $0$, since $X(0)+\tau I=\mathrm{diag}(1+\tau ,\tau )⪰0$ for $\tau \ge 0$ and is impossible for $\tau <0$ by testing the second coordinate.

Every matrix in the slice $X_{22}=0$, $X_{12}=1$ has form $X(t)=\left(\begin{array}{cc}t& 1\\ 1& 0\end{array}\right)$. If $t>0$, the vector $(-1\mathrm{/}t,1)$ gives quadratic form $-1\mathrm{/}t<0$; if $t=0$, vector $(-1,1)$ gives $-2<0$; if $t<0$, vector $(1,0)$ gives $t<0$. Thus the slice is infeasible. For $t>0$, the matrix $Y(t)=\left(\begin{array}{cc}t& 1\\ 1& 1\mathrm{/}t\end{array}\right)$ is positive semidefinite because its quadratic form is $t(u+v\mathrm{/}t{)}^2$; its Frobenius distance from $X(t)$ is $1\mathrm{/}t\to 0$. For any $\tau >0$, choose $t=1\mathrm{/}\tau$. Then $X(t)+\tau I$ has quadratic form $(\frac{1}{\tau }+\tau )u^2+2uv+\tau v^2=(\frac{1}{\tau }+\tau ){(u+\frac{\tau v}{1+{\tau }^2})}^2+\frac{{\tau }^3}{1+{\tau }^2}{v}^2$, which is positive for every nonzero $(u,v)$, so it is positive definite and $\tau$ is Phase-I feasible. If $\tau <0$, the lower-right diagonal entry of $X(t)+\tau I$ is $\tau <0$, so it cannot be positive semidefinite. Hence the Phase-I infimum is zero and is not attained.