  We prove the three embeddedness results as follows. $({\rm i})$ Let
$\Gamma_{2m+1}$ be a piecewise geodesic Jordan curve with $2m+1$ vertices in
$\mathbb{R}^n$, where $m$ is an integer $\geq2$. Then the total curvature of
$\Gamma_{2m+1}<2m\pi$. In particular, the total curvature of $\Gamma_5<4\pi$
and thus any minimal surface $\Sigma \subset \mathbb{R}^n$ bounded by
$\Gamma_5$ is embedded. Let $\Gamma_5$ be a piecewise geodesic Jordan curve
with $5$ vertices in $\mathbb{H}^n$. Then any minimal surface $\Sigma \subset
\mathbb{H}^n$ bounded by $\Gamma_5$ is embedded. If $\Gamma_5$ is in a geodesic
ball of radius $\frac{\pi}{4}$ in $\mathbb{S}^n_+$, then $\Sigma \subset
\mathbb{S}^n_+$ is also embedded. As a consequence, $\Gamma_5$ is an unknot in
$\mathbb{R}^3$, $\mathbb{H}^3$ and $\mathbb{S}^3_+$. $({\rm ii})$ Let $\Sigma$
be an $m$-dimensional proper minimal submanifold in $\mathbb{H}^n$ with the
ideal boundary $\partial_{\infty} \Sigma = \Gamma$ in the infinite sphere
$\mathbb{S}^{n-1}=\partial_\infty \mathbb{H}^n$. If the M{\"o}bius volume of
$\Gamma$ $\widetilde{\vol}(\Gamma) < 2\vol(\mathbb{S}^{m-1})$, then $\Sigma$ is
embedded. If $\widetilde{\vol}(\Gamma) = 2\vol(\mathbb{S}^{m-1})$, then
$\Sigma$ is embedded unless it is a cone. $({\rm iii})$ Let $\Sigma$ be a
proper minimal surface in $\hr$. If $\Sigma$ is vertically regular at infinity
and has two ends, then $\Sigma$ is embedded.
