A "long topological knot" in $\mathbb R^n$ is a topological embedding $f : \mathbb R \to \mathbb R^n$ such that $f(x) = (x,0)$ for all $x \in \mathbb R \setminus (-1,1)$.

Let $K_n$ be the space of all long topological knots in $\mathbb R^n$ with the compact-open topology. Then $K_n$ is contractible. The contraction is given by

$F : [0,1] \times K_n \to K_n$ defined by

$F(t,f)(x) = (1-t)f(\frac{x}{1-t})$ provided $t \in [0,1)$ and $F(1,f)(x) = (x,0)$.

This map, $F$, is sometimes called "The Alexander Trick". See the Wikipedia Alexander Trick page for context.

In response to your edit, perhaps an interesting topology on $K_n$ could be given this way. Given $f \in K_n$ and $\epsilon > 0$ we'll say an $\epsilon$-ball about $f$ consists of all knots $\phi \circ f$ where $\phi : \mathbb R^n \to \mathbb R^n$ is a homeomorphism which agrees with the identity map outside of $D^n$, and such that $|\phi(x)-x|<\epsilon$ for all $x \in \mathbb R^n$. The topology on $K_n$ could be the topology generated by all $\epsilon$-balls about all $f \in K_n$. Presumably this kind of topology has a name?

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A "long topological knot" in $\mathbb R^n$ is a topological embedding $f : \mathbb R \to \mathbb R^n$ such that $f(x) = (x,0)$ for all $x \in \mathbb R \setminus (-1,1)$.

Let $K_n$ be the space of all long topological knots in $\mathbb R^n$ with the compact-open topology. Then $K_n$ is contractible. The contraction is given by

$F : [0,1] \times K_n \to K_n$ defined by

$F(t,f)(x) = (1-t)f(\frac{x}{1-t})$ provided $t \in [0,1)$ and $F(1,f)(x) = (x,0)$.

This map, $F$, is sometimes called "The Alexander Trick". See the Wikipedia Alexander Trick page for context.