Condition for image to pertain to the set Suppose we have a continous closed function $f$ between $[0,1]$ and $\mathbb{R}^2$. Suppose $Y_1$ is a subset of $\mathbb{R}^2$ and that  $f(1) \in Y_1$.
Can we assure that $t_0 = \inf \{ t \in [0,1] : f(t) \in Y_1 \}$ satisfies $f(t_0) \in Y_1$? In case we can't, which conditions are required? Is compactness of $Y_1$ sufficient? I am having problems translating topology notions into analytic concepts.
 A: If $Y_1$ is closed, your $t_0$ is just $\inf f^{-1}[Y_1]$ which is closed in $[0,1]$ and so contains its infimum. This implies $f(t_0) \in Y_1$ indeed.
We only need $f$ to be continuous, $Y_1$ closed in $\Bbb R^2$ for this argument to work. For the domain $[0,1]$ we use its closedness in $\Bbb R$, essentially (and the fact that $f^{-1}[Y_1]$ is bounded below).
No closedness of $f$ required (which in your case is automatic by compactness of $[0,1]$). For non-closed $Y_1$ this could easy fail.
A: No, you cannot guarantee it with no assumptions on $Y$. For example, take $f\colon [0,1]\to \mathbb{R}^2$ given by $f(a) = (a,0)$, and let $Y=(0,2)\times\mathbb{R}$. Then $f(1)\in Y$, and $\{a\in [0,1]\mid f(a)\in Y\} = (0,1]$, so $t_0=0$ and $f(t_0)\notin Y$.
You can ensure that $f(t_0)\in Y$ if $Y$ is closed, since $t_0$ is the limit of a sequence of elements of $\{a\in[0,1]\mid f(a)\in Y\}$, so by continuity $f(t_0)$ is a limit point of $Y$. This does not show the condition is "required", but it is certainly sufficient.
The condition, though, is in fact necessary. Let $Y$ be any set that is not closed; let $(y_n)_{n\in\mathbb{N}}$ be a sequence of points of $Y$ that converge to a point $y_0\notin Y$. Define $f\colon [0,1]\to \mathbb{R}^2$ by a piecewise linear function that sends $[\frac{1}{n+1},\frac{1}{n}]$ to the line from $y_{n+1}$ to $y_n$, and sends $0$ to $y_0$. It is not hard to check that this function is continuous, $f(1)\in Y$, but $t_0=0$ and $f(t_0)=y_0\notin Y$.
