The following theorem is implicit in the paper by Blankinship, where it is proven that for each $n\ge 3$ and $q\in [1, n]$, there exists a subset $A\subset {\mathbb R}^n$ homeomorphic to the (closed) $q$-dimensional cube and whose complement is not simply-connected. Note that this result also shows that not all 2-dimensional topological disks in ${\mathbb R}^4$ are isotopic to each other.

*William A. Blankinship*, **Generalization of a construction of Antoine**, Ann. Math. (2) 53, 276-297 (1951). ZBL0042.17601.

**Theorem.** For every $n\ge 4$ there exists a nontrivial (wild) 1-dimensional knot $K\subset {\mathbb R}^n$.

Proof. Blankinship proves that for each $n\ge 4$, there exists a subset $C\subset R^n$ homeomorphic to the classical cantor set such that $R^n\setminus C$ is not simply-connected (the complement is necessarily connected by the Alexander duality). Now, take such $C$. I proved in my answer here that for every totally disconnected compact $C$ in ${\mathbb R}^n$ (e.g. one as above), there exists a subset $K\subset {\mathbb R}^n$ homeomorphic to $S^1$ and containing $C$. Moreover, in the construction $K$ can be chosen such that each component of $K\setminus C$ is a smooth (open) arc in ${\mathbb R}^n$. (There are at most countably many of these components.)

**Lemma.** $\pi_1({\mathbb R}^n \setminus K)\ne 1$.

Proof. Since $\pi_1({\mathbb R}^n \setminus C)\ne 1$, there is a smooth loop $c: S^1\to {\mathbb R}^n \setminus C$ which is not null-homotopic in ${\mathbb R}^n \setminus C$. A priori, the image of $c$ intersects $A=K \setminus C$. However, by perturbing $c$ to be transversal to $A$, since $n\ge 2$, we can find a loop $c_1$ homotopic to $c$ in $R^n \setminus K$ and whose image is disjoint from $A$. Hence, $c_1$ is a homotopically nontrivial loop in ${\mathbb R}^n
\setminus K$ (since it is already homotopically nontrivial in the complement to $C$). qed

On the other hand, if $T\subset {\mathbb R}^n$, $n\ge 4$, is a trivial knot, $\pi_1({\mathbb R}^n\setminus T)=1$. Thus, the knot $K$ constructed above nontrivial. qed

What Andrew's answer proves that every **tame** 1-dimensional knot in ${\mathbb R}^4$ (and, more generally, ${\mathbb R}^n, n\ge 4$) is trivial, i.e. is ambient isotopic to a round circle. The knot constructed in my proof is **wild.**

In particular, step 1 in Andrew's answer breaks down for general knots.

On the positive side:

**Theorem.** If $K\subset {\mathbb R}^3$ is a topological knot, then $K$ is unknotted in ${\mathbb R}^4$.

**Proof.** The key is to prove that $K$ is tame (or *locally flat*) as a knot in ${\mathbb R}^4$, i.e. for every $x\in K$ there is a neighborhood $U$ of $x$ in ${\mathbb R}^4$ and a homeomorphism $U\to V\subset {\mathbb R}^4$ sending $U\cap K$ to a subset of a straight line $L\subset {\mathbb R}^4$.

Once it is done, it follows from the main theorem of

H. Gluck, **Unknotting $S^ 1$ in $S^ 4$**, Bull. Am. Math. Soc. 69, 91-94 (1963). ZBL0108.36503.

that $K$ is unknotted in ${\mathbb R}^4$.

Let $f: K\to [0,1]$ with exactly two point of minimum and maximum, $f(a)=0, f(b)=1$. Let $\alpha, \beta$ denote the closures of the components of $K\setminus \{a, b\}$. The function $f$ defines an embedding $K\to {\mathbb R}^4$,
$$
\phi: x\mapsto (x, f(x)), x\in K,
$$
whose image is a knot $K'$ in ${\mathbb R}^4$. Let $a':= \phi(a)=a, b':= \phi(b)$.

Step 1. $K'$ is ambient-isotopic to $K$ in ${\mathbb R}^4$.

First, extend $f$ to a continuous function $f: {\mathbb R}^3\to [0,1]$.
For each $t\in [0,1]$ define the "shear"
$$
h_t(x, s)= (x, s+tf(x)), x\in {\mathbb R}^3, s\in {\mathbb R}
$$
which is clearly a homeomorphism. This defines an ambient isotopy of the knot $K$ to the knot $K'=f_1(K)$ in ${\mathbb R}^4$.

Step 2. I will now prove a slightly easier result than unknottedness of $K'$ in ${\mathbb R}^4$, I will show that the arc $\alpha'=\phi(\alpha)$ (and, similarly, $\beta'=\phi(\beta))$ is tame, equivalently, unknotted in $R^4$. Note that $\alpha'$ intersects each horizontal slice ${\mathbb R}^3 \times \{s\}\subset {\mathbb R}^4$ in exactly one point, namely,
$(f^{-1}(s), s)$. The straight-line segment $I':=a'b'$ intersects each slice ${\mathbb R}^3 \times \{s\}\subset {\mathbb R}^4$ also at exactly one point, $p_s$.
Thus, the shear
$$
h^\alpha: (x, s) \mapsto (x + (p_s- f^{-1}(s)), s), s\in [0,1]
$$
is a homeomorphism ${\mathbb R}^3\times [0,1]\to {\mathbb R}^3\times [0,1]$ sending $\alpha'$ to $I'$; this homeomorphism is the identity on the boundary of the slab ${\mathbb R}^3\times [0,1]$. Hence, $h$ extends to a homeomorphism ${\mathbb R}^4\to{\mathbb R}^4$ (by the identity outside of the slab). Linear interpolation again yields an isotopy $h^\alpha_t, t\in [0,1]$, of $h$ to the identity map. Hence, the arc $\alpha'$ is unknotted in ${\mathbb R}^4$. Note that $\{h^\alpha_t(\alpha'): t\in [0,1]\}$ lies inside the convex hull of the arc $\alpha'$ in ${\mathbb R}^4$.

The same construction applies to the arc $\beta'$ resulting in an isotopy $h^\beta$. The trouble is that we cannot combine these "straightening" maps of $\alpha', \beta'$. We obtain, however, that the knot $K'$ is tame in ${\mathbb R}^4$, except possibly at the points $a', b'$. More precisely, $K'$ is the concatenation of two tame arcs, $\alpha', \beta'$. Since $K$ is ambient-isotopic to $K'$ in ${\mathbb R}^4$, it follows that $K$ was tame in ${\mathbb R}^4$ as well, except possibly at the points $a, b$. But the points $a, b$ in $K$ were chosen arbitrarily. Hence, $K\subset {\mathbb R}^4$ locally flat at every point, i.e. is tame.