First, I wasn't exactly sure which question I should ask, for the main title, but hopefully this explanation helps.
It appears to be well known that the complement of a knot in $\mathbb{R}^3$ is always path connected. But it seems to me that a space-filling knot could exist, which would make the ...
I may be wrong, but I don't see how a continuous map $K:[0,1]\to\mathbb{R}^3$ could satisfy $K([0,1])=\mathbb{R}^3$. The interval $[0,1]$ is compact but $\mathbb{R}^3$ is not.
@MichaelMorrow that's not one of the constraints, the knot $K$ just has to satisfy $K(0)=K(1)$ and, in this specific case, $\mathbb{R}^2\subseteq K([0,1])$.
Because the continuous image of a compact set must be compact. So, since $[0,1]$ is compact, whatever $K([0,1])$ is, it needs to be compact. Which, since $\mathbb{R}^3$ is a metric space, means it needs to be bounded, hence cannot fill space.
Let $X\subset \mathbb R^{n}$ be a compact set, and $f :\mathbb R^{n}\to \mathbb R $ a continuous function. Then, $F(X)$ is a compact set.
I know that this question may be a duplicate, but the problem is that I have to prove this using real analysis instead of topology.
I'm struggling with prov...
Let $X\subset \mathbb R^{n}$ be a compact set, and $f :\mathbb R^{n}\to \mathbb R $ a continuous function. Then, $F(X)$ is a compact set.
I know that this question may be a duplicate, but the problem is that I have to prove this using real analysis instead of topology.
I'm struggling with prov...
Discussion between Calvin Godfrey and…
Discussion between Calvin Godfrey and…