Embed the Cayley graph of the free group $F_2$ in $\mathbb{R}^2 \subset \mathbb{R}^3$ (note that the edges must shrink geometrically to avoid self-intersection). Increase the vertical component of each vertex $v$ by an amount $h_v$ to be described below. This is your isotopy at $t=0$. Now, as $t \to 1$, stretch the edges to unit length and scale the vertical height of vertices by $(1-t)h_v$, so they full thing comes to rest back on the plane. In the end you covered the infinite square grid that represents the Cayley graph of $\mathbb{Z} \times \mathbb{Z}$. To ensure that this gives an embedding for every $t$, write $v = x_1 \ldots x_r$ with $x_i \in \{ a,b,a^{-1},b^{-1} \}$ (generators of $F_2$), and let $h_v = \sum_{x_j \in \{a,a^{-1}\}} 2^{-j}$. In other words, read $x_1, \ldots, x_r$ as a binary fraction replacing $a,a^{-1}$ by 1 and $b,b^{-1}$ by 0. This way any pair of edges that run the risk of overlapping will have disjoint heights and not intersect after all.

For a second example, take the standard Cantor set in $[-1,1]$. It consists of two copies of itself scaled down by $1/3$ in $\big[-1,-\frac13\big] \cup \big[\frac13,1\big]$. As $t \to 1$, rotate $[-1,0]$ by $\pi t$ about 0, while rescaling by $3t$ and translating the fold point to $-t$. The Cantor set covered itself !

Embed the Cayley graph of the free group $F_2$ in $\mathbb{R}^2 \subset \mathbb{R}^3$ (note that the edges must shrink geometrically to avoid self-intersection). Increase the vertical component of each vertex $v$ by an amount $h_v$ to be described below. This is your isotopy at $t=0$. Now, as $t \to 1$, stretch the edges to unit length and scale the vertical height of vertices by $(1-t)h_v$ so the full thing comes to rest back on the plane. In the end you covered the infinite square grid that represents the Cayley graph of $\mathbb{Z} \times \mathbb{Z}$. To ensure that this gives an embedding for every $t$, write $v = x_1 \ldots x_r$ with $x_i \in \{ a,b,a^{-1},b^{-1} \}$ (generators of $F_2$), and let $h_v = \sum_{x_j \in \{a,a^{-1}\}} 2^{-j}$. In other words, read $x_1, \ldots, x_r$ as a binary fraction replacing $a,a^{-1}$ by 1 and $b,b^{-1}$ by 0. This way any pair of edges that run the risk of overlapping will have disjoint heights and not intersect after all.

For a second example, take the standard Cantor set in $[-1,1]$. It consists of two copies of itself scaled down by $1/3$ in $\big[-1,-\frac13\big] \cup \big[\frac13,1\big]$. As $t \to 1$, rotate $[-1,0]$ by $\pi t$ about 0, while rescaling by $3t$ and translating the fold point to $-t$. The Cantor set covered itself !

Embed the Cayley graph of the free group $F_2$ in $\mathbb{R}^2 \subset \mathbb{R}^3$ (note that the edges must shrink geometrically to avoid self-intersection). Increase the vertical component of each vertex by $n$, where $n$ is its distance to the identity. This is your isotopy at $t=0$. Now, as $t \to 1$, stretch the edges to unit length and scale the vertical height of vertices by $(1-t)$, so they full thing comes to rest back on the plane. In the end you covered the infinite square grid that represents the Cayley graph of $\mathbb{Z} \times \mathbb{Z}$.

For a second example, take the standard Cantor set in $[-1,1]$. It consists of two copies of itself scaled down by $1/3$ in $\big[-1,-\frac13\big] \cup \big[\frac13,1\big]$. As $t \to 1$, rotate $[-1,0]$ by $\pi t$ about 0, while rescaling by $3t$ and translating the fold point to $-t$. The Cantor set covered itself !

Embed the Cayley graph of the free group $F_2$ in $\mathbb{R}^2 \subset \mathbb{R}^3$ (note that the edges must shrink geometrically to avoid self-intersection). Increase the vertical component of each vertex $v$ by an amount $h_v$ to be described below. This is your isotopy at $t=0$. Now, as $t \to 1$, stretch the edges to unit length and scale the vertical height of vertices by $(1-t)h_v$, so they full thing comes to rest back on the plane. In the end you covered the infinite square grid that represents the Cayley graph of $\mathbb{Z} \times \mathbb{Z}$. To ensure that this gives an embedding for every $t$, write $v = x_1 \ldots x_r$ with $x_i \in \{ a,b,a^{-1},b^{-1} \}$ (generators of $F_2$), and let $h_v = \sum_{x_j \in \{a,a^{-1}\}} 2^{-j}$. In other words, read $x_1, \ldots, x_r$ as a binary fraction replacing $a,a^{-1}$ by 1 and $b,b^{-1}$ by 0. This way any pair of edges that run the risk of overlapping will have disjoint heights and not intersect after all.

For a second example, take the standard Cantor set in $[-1,1]$. It consists of two copies of itself scaled down by $1/3$ in $\big[-1,-\frac13\big] \cup \big[\frac13,1\big]$. As $t \to 1$, rotate $[-1,0]$ by $\pi t$ about 0, while rescaling by $3t$ and translating the fold point to $-t$. The Cantor set covered itself !