Let $X$ be a manifold with a boundary. That is, for any $x\in X$ there is a neighborhood about $x$ diffeomorphic to an open subset of $H^k=\{(x_1,x_2,\ldots,x_k)\in R^k: x_k \geq 0\}$. The boundary of $X$, denoted, $\partial X$, is the set of points whose image is has the last coordinate as zero. The interior of $X$, $Int(X)$, is the compliment of $\partial X$ in $X$.
Show that $Int(X)$ is open in $X$.
Attempt: Let $x\in Int(X)$. Then there is a chart $(U,\varphi)$, $U$ is a neighborhood of $x$ which is mapped diffeomorphically to $\varphi(U)\subset R^k$. We want to show that there is some open set $W$ containing $x$, such that $W\subset Int(X)$. Pick any $W\subset U$, then $\varphi(W)$ is an open set in $R^k$. Therefore $W\subset Int(X)$.
Best Answer
$\varphi(U)$ is not necessarily open in $\mathbb{R}^k$, only in $\mathbb{H}^k$.
However, we know that $\varphi(x) \in \operatorname{Int} \mathbb{H}^k = \{(x_1, \ldots, x_n) \in \mathbb{H}^k : x_n > 0\}$ so there exists an open neighbourhood $V \subseteq U$ of $x$ such that $\varphi(V) \subseteq \operatorname{Int} \mathbb{H}^k$ (namely, take $V = U \cap \varphi^{-1}(\operatorname{Int} \mathbb{H}^k)$).
Since $\varphi$ is a diffeomorphism, $\varphi(V)$ is open in $\operatorname{Int} \mathbb{H}^k$ and hence open in $\mathbb{R}^k$.
For every $y \in V$ we have that $(V, \varphi|_V)$ is a chart around $y$ such that $\varphi(y) \in \operatorname{Int} \mathbb{H}^k$ so $y \in \operatorname{Int} X$. Hence $V \subseteq \operatorname{Int} X$.