Sets open with respect to [0,1]

Question

For a fucnton defined as $f:(0,1)\rightarrow \mathbb{R}$ (Continuous and strictly increasing), we also have $f^{'}:[0,1]\rightarrow \mathbb{R}\cup(-\infty,\infty).$ Why is the set $f^{-1}((-\infty,d))=[0,f^{-1}(d))$ open in $[0,1]?$

Because from how I understand open sets, for some open set $U$ in $X,$ how can $U=[0,f^{-1}(d))$ be open in $[0,1]?$ For instance if $0\in U$, then we cant always find some open ball $B(0,\epsilon)\subset U,$ cause then it will lie outside of $0,$ and that portion is no longer in $U$, so what am I missing here?

Answer 1

[0,a) is open within [0,1] because
[0,a) = (-a,a) $\cap$ [0,1] is in the relative topology.

In otherwords within [0,1],
[0,a) = B(0,a) = { x in [0,1] : |0 - x| < a }

Answer 2

This is a basic question related to the subspace of a topological space. If $(X,\mathcal{T}_{X})$ is a topological space and $Y \subset X$, then the induced topology on $Y$ due to $X$ is given by:

$$ \mathcal{T}_{Y} = \{ Y \cap B : B \in \mathcal{T}_{x} \} $$

So, the topological space induced by $\mathbb{R}$ with euclidean topology on $[0,1]$ will have open sets of the type:

$\{ (a,b): 0 \leq a<b \leq 1 \} \cup \{ (c,1]: 0 \leq c < 1 \} \cup \{ [0,d): 0 < d \leq 1 \} \cup \{ [0,1] \} $