Continuous map between subsets of topological spaces I know very little about topology, so this is a rather basic question.
A continuous map between topological spaces $X$ and $Y$ is defined as a function $f\colon X\to Y$ such that the preimage of any open set is itself an open set. This generalises the idea of, say, a continuous function from $\mathbb{R}$ to $\mathbb{R}$, as it would be defined in real analysis.
However, in analysis one would often want to say that a function is continuous on an interval. For example, I might have a continuous function $f\colon [0,1]\to\mathbb{R}$. How would one talk about such a function in the language of topological spaces?
If the domain was an open interval I could guess the answer: I could define a new topological space consisting of the interval and all its open subsets, and then a continuous function $g\colon (0,1)\to\mathbb{R}$ would just be a continuous map between the topological spaces $(0,1)$ and $\mathbb{R}$ (with their usual topology).
This doesn't seem to work for $f\colon [0,1]\to\mathbb{R}$, because $[0,1]$ isn't a topological space. (At least not in an obvious way.) So how does one talk about continuous functions in this kind of context?
 A: $[0, 1]$ is a topological space, given something called the subspace topology. We say that the open sets of $[0, 1]$ are precisely the sets of the form $U \cap [0, 1],$ where $U$ is an open subset of $\mathbb{R}.$
This can lead to some counter intuitive behavior at first: $[0,1]$ is an open subset of $[0, 1],$ even though it isn't in $\mathbb{R}.$ The set $[0, 0.17)$ is now open as well.
Why this definition, though? Well, this definition has a very nice defining property. Let's suppose we have a topological space $X,$ and a subset $S\subseteq X.$ What topology should we give $S$?
Well, it should make it so that the inclusion map $i : S\rightarrow X$ that just takes a point in $S$ and 'remembers' it belongs to $X$ continuous. To make this continuous, you need so that if $U\subseteq X$ is open in $X,$ then $i^{-1}(U) = U\cap S$ to be open in $S.$ So, if you want the inclusion map to be continuous, the easiest thing to do is to define the topology to be so that the open sets are of the form $U\cap S.$ This way you get continuity, while not doing anything more than the bare minimum to ensure continuity.
A: The standard topology for a subspace $A$ of a topological space $X$, is the subspace topology, or relative topology, defined by
$\tau_A=\{U\subseteq A~~| U=A\cap V~~\text{for}~~ V\subseteq X~~\text{open}\}$.
You can check easily that this defined indeed a topology on $A$.
For example $(0,1)$ is open in $[0,1]$ with regards to this topology, as $(0,1)$ is open in $\mathbb{R}$ and $(0,1)\cap [0,1]=(0,1)$.
Also $[0,1]$ is open with regards to this topology, as $[0,1]\cap\mathbb{R}=[0,1]$.
This is 'surprising' as $[0,1]$ is not open in $\mathbb{R}$. Also a set like $(1/2,1]$ is open in $[0,1]$. Why?
So the open sets in the subspace topology can look quite different.
As I said the subspace topology is the usual topology when we look at subsets. But of course you can have other topologies, but for a subset this is the 'best' one, as it has a so called universal property.
A: 
This doesn't seem to work for $f\colon [0,1]\to\mathbb{R}$, because $[0,1]$ isn't a topological space. (At least not in an obvious way.) So how does one talk about continuous functions in this kind of context?

Consider $I_\mathfrak{U}=[0,1]_\mathfrak{U}$ , in the usual subspace topology inherited from $\Bbb R_\mathfrak{U}.$
As you are new in General topology so It might be little confusing for you.
$[0,1]$ might be open or closed in subspace topology inherited from usual topology on $\Bbb{R}$.
So one can prove continuity $f:[0,1]\to \Bbb{R}.$
For example: function $f:[0,1]\to \Bbb R^2$ defined by $f(t)=(cos2\pi t,sin2\pi t)$.
This is a good example to show $f$ is bijective continuous but not bicontinuous.
Bicontinuous stands for both $f$ and $f^{-1}$ are continuous.
