Can $S^k$ be embedded in $S^n$ for all $k>n$? In Hatcher's Algebraic Topology, Section 2.B, there is a proposition concerning about computing the homology of $S^n-h(S^k)$, where $k<n$ and $h:S^k \to S^n$ is an embedding. It is clear that there exists an embedding $S^k \to S^n$ for $k\leq n$. 
What I'm curious is, can $S^k$ be embedded in $S^n$ for $k>n$? Intuitively this seems not true, but is there a way to show this?
Also I'm curious for a similar question: Can the closed disk $D^k$ be embedded to $S^n$ for $k>n$?
Thanks in advance.
 A: The answer is no: this contradicts topological invariance of dimension. This is Theorem 2.26 of Hatcher:

If nonempty open subsets $U\subseteq \mathbb{R}^m$ and $V\subseteq \mathbb{R}^n$ are homeomorphic, then $m=n$. 

If $k>n$, then an embedding of $S^k$ into $S^n$ induces (using Stereographic projection) a homeomorphism of an open subset of $\mathbb{R}^k$ onto an open subset of $\mathbb{R}^n$. This contradicts the topological invariance of dimension theorem.
The same deal applies for the closed disk. An embedding of $D^k$ to $S^n$ restricts to a homeomorphism of an open subset of $\mathbb{R}^k$ onto an open subset of $\mathbb{R}^n$ again using stereographic projection. This contradicts topological invariance of dimension. 
A: This is a consequence of dimension theory: $\dim(S^n)=n$ for all $n$ and for separable metric spaces $A \subseteq X$ implies $\dim(A) \le \dim(X)$ (this is easy for small inductive dimension, and all standard dimension functions coincide for sep. metric spaces). So $S^k$ can never embed into $S^n$ when $k > n$. 
Invariance of dimension (used in another answer) is overkill here.
