$X$ non-degenerate Hausdorff space , $f:[0,1] \to X$ is surjective continuous , $a:=\inf \{y\in (0,1]: f([0,y])=X\}$ , then is $a>0$ and $f([0,a])=X$?

Question

Let $X$ be a Hausdorff space with at least two points and $f:[0,1] \to X$ be a surjective continuous map ; let $a:=\inf \{y\in (0,1]: f([0,y])=X\}$ , then is it true that $a>0$ and $f([0,a])=X$ ?

Answer 1

Suppose $f[0,a]\neq X$ (in particular $a<1$) and take $y\in X\setminus f[0,a]$. Take an open set $U$ in $X$ containing $f[0,a]$ but not $y$ (by the Hausdorff property). Then $f^{-1}(U)$ is open in $[0,1]$ and contains $[0,a]$. But if $t\geq a$ with $t\in f^{-1}(U)$, then $f([0,t])\subseteq f(f^{-1}(U))\subseteq X\setminus y$. This contradicts the definition of $a$ as an infimum $\left.\right.^{(1)}$.

Therefore $f[0,a]=X$. In particular $a>0$ because $X$ has more than one point (and $f[0,0]=\{f(0)\}$).

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(1) A little more details on why this contradicts the definition of $a$: there exists $\epsilon>0$ such that $(a-\epsilon,a+\epsilon)\subseteq f^{-1}(U)$. The rest of the argument shows that $[a,a+\epsilon)\cap\left\{y\in[0,1]:f[0,y]=X\right\}=\varnothing$, and this is not possible since $a$ is the infimum of $\{y\in[0,1]:f[0,y]=X\}$.

Answer 2

Yes.

Suppose that $f\big[[0,a]\big]\ne X$, and fix $x\in X\setminus f\big[[0,a]\big]$. For each $y\in(a,1]$ there is a $z_y\in(a,y]$ such that $f(z_y)=x$. Let

$$Z=\{z_y:y\in(a,1]\}\;;$$

clearly $a\in\operatorname{cl}Z$, so $f(a)=x$. (It’s clear that $a>0$, since $|X|>1$.)