Continuous injective curve from $[0,1]$ to $\mathbb{R^2}$ containing the interval $[0,1]\times {0}$ I want to know if it is possible to have an injective continuous curve from $[0,1]$ to $\mathbb{R^2}$ whose image contains the interval $[0,1]\times \{0\}$ with $f(0)=(0,0)$. There is one trivial solutions, that is, go straight across the interval: $f(x)= (x,0)$. Let $f(t_f)=(1,0)$. We say two solutions $f,g$ are different if $f([0,t_f])\neq g([0,t_g])$.
Does there exist an example different from the trivial solution? i.e. the curve leaves the interval at some point before finishing.
If yes, please give a counter example. Otherwise, please put the solution in spoiler tags, I have been trying to solve this myself but I'm too worried there is a counter example.
My current idea is to look at the sup $t_0=\{t: \exists s: f([0,t])= [0,s]\times \{0\} \}$. Intuitively, to cover $(t_0+\epsilon,0)$, we have to come back later. We find a sequence of epsilons converging to zero such that the time of return does not converge to $0$. Then by continuity (+Bolzano weierstrass on epsilons) we get $f(\lim \epsilon_n)=(0,0)$ contradicting injectivity. Im having a hard time materialising the third sentence.
edit:
My solution:
Suppose $f$ is different from the trivial solution. Then we must have some $t'<t_f$ where we are not on the interval and so by continuity we can have an interval. Lets take a maximal one i.e. let
$$t_0 := \inf \{t: \pi(f(t))\neq 0, t\le t' \} $$
$$t_1 := \inf \{t: \pi(f(t))=0 \cap t\ge t'\}$$
Then by continuity,

*

*$t_0\neq t_1$

*$\pi_y(f(t_0)) = \pi_y(f(t_1))=0$ but $\pi_y(f(t)) \neq 0 $ for $t_0<t<t_1$
and by injectivity

*

*$\pi_x(f(t_0))\neq \pi_x(f(t_1))$
Some of the interval $(\pi_x(f(t_0)), \pi_x(f(t_1)))\times \{0\}$ may have been covered in $[0,t_0]$, however, not all of it otherwise $f(t_1)$ is an intersection. What was covered is, by continuity, a closed set therefore the complement in the interval is $(\text{open set in }\mathbb{R}) \times \{0\} $ in particular we have a "maximal" subset  $(a, \pi_x(f(t_1) ))\times \{0\}$ for which $(a,0)\in f([0,t_0])$. $(a, \pi_x(f(t_1) ))\times \{0\}$ must be covered in $f[t_1,1]$ therefore by continuity $\exists t_2\in(t_1,1]$ with $f(t_2)=(a,0)$, contradicting injectivity
 A: It is not possible, and here is a proof (don't look if you don't want to see!)
We recall the following facts about topological spaces:

*

*The image of a compact set under a continuous function is compact

*A compact subset of Euclidean space is closed and bounded (Heine-Borel Theorem)

*In a connected topological space $X$, the only open and closed sets are $\emptyset$ and $X$.

Now, let's say we have a function $g:[0,1]\rightarrow \mathbb{R}^2$ different from the trivial example whose image contains the required interval. By your definition, there is $s \in [0,1]$ such that $g(s) \not\in [0,1]\times\{0\}$. Since $g$ is continuous, there exists $\varepsilon > 0$ such that $g([s-\varepsilon, s+\varepsilon]) \cap ([0,1]\times \{0\}) = \emptyset$.
Moreover, $[s+\varepsilon,t_g]$ is compact, so $g([s+\varepsilon,t_g])$ is also compact, and therefore closed and bounded. $g([s+\varepsilon,t_g])\cap([0,1]\times\{0\})$ is therefore a closed subset of $[0,1]\times\{0\}$ (using the subspace topology). The subset $g([0,s-\varepsilon])\cap([0,1]\times\{0\})$ is also a closed subset of $[0,1]\times\{0\}$ by the same argument. Since the union of these two sets is the entirety of $[0,1]\times \{0\}$ by the previous paragraph, both are also the complement of closed sets and therefore open sets.
Since these two subsets are open and closed, and nonempty, they must both be the entire space $[0,1]\times\{0\}$. They are therefore not disjoint, and the map $g$ is not injective.
A: The answer is negative, as I understand your question. If you start at $f(0) = (0, 0)$, want $f$ to be continuous and that $f([0, 1])$ contains $[0, 1] \times \{0 \}$, you have to start by tracing this line.
Indeed, consider : $\{ t \in [0, 1], \pi_y(f(t)) = 0 \}$ where $\pi_y$ denotes the projection on the second coordinate. By continuity of $f$, this is a closed set and therefore a compact subset of $[0, 1]$. Each connected component of this compact is itself compact, and there are at most countably many. Now the image by $f$ of each such compact connected component is a compact interval by continuity.
By projecting on the $x$-coordinate, this means that you can write $[0, 1]$ as the countable disjoint (by injectivity of $f$) union of compact intervals. But this is impossible, unless there is exactly one such compact interval. (At least with the axiom of choice.) This mean that $\{ t \in [0, 1], \pi_y(f(t)) = 0 \}$ is an interval, and since it contains $0$ you are done.
A: One counterexample to your claim is the curve which traces out the unit square except for one side. This curve would be defined piecewise, by
$$
f(t) = \begin{cases}(3t,0)&0\leq t < 1/3 \\
(1,3(t-1/3)&1/3\leq t<2/3 \\
(1-3(t-2/3), 1)&2/3\leq t \leq 1 \end{cases}
$$
Whose image looks like this:

Clearly, this curve contains $[0,1]\times[0]$ and is not the trivial solution you provided which contains only the interval. It is also injective, as every point in the image curve is associated with a unique domain point $t\in[0,1]$.
