With the usual metric in $\mathbb{R}$, there is no open interval around $0$ that is completely contained in $[0,1/2)$. However, I have come across the fact that "$[0,1/2)$ is open in $[0,1]$" in a proof I am going through. It seems to directly contradict the definition of openness. A rationale behind this would be appreciated.
[Math] Why is [0,1/2) open in [0, 1]
general-topologymetric-spaces
Related Solutions
Ask yourself if the spaces $C^k(0,1)$ are of interest.
EDIT: I also suggest thinking more about and therefore formalising the notion of a 'good' metric on a space. You might find that a good metric is one that induces a topology on the space that has desirable properties for doing analysis, like local convexity, being complete, the Heine-Borel property, an interesting dual space, etc.
Since $(0,1)$ is the union of countably many compact sets $K_n$ such that $K_n\subset\text{int}K_{n+1}$, then we define $C(0,1)$ as the space of all continuous real-valued functions on $(0,1)$ with topology induced by the family of separating semi-norms: $$ p_n(f):=\sup_{x\in K_n}|f(x)|. $$ Then the sets $$ V_n:=\{f\in C(0,1)\;|\;p_n(f)<1/n\},\qquad n=1,2,... $$ form a local convex base for $C(0,1)$. This topology is in fact induced by the metric $$ d(f,g):=\max_n\frac{1}{2^n}\frac{p_n(f-g)}{1+p_n(f-g)} $$ This metric is complete, so $C(0,1)$ becomes a Frechet space.
The space $C^\infty(0,1)$ is defined to be the space of all real-valued functions $f$ on $(0,1)$ with the property that $D^kf\in C(0,1)$ for all $k\in\mathbb{N}_0$. Choosing the same compact sets $K_n$ as above, the following family of semi-norms undices a metrizable, locally-convex topology on $C^\infty(0,1)$: $$ p_n(f):=\max_{x\in K_n,\;k\leq n}|D^kf(x)|. $$ The metric on this space is of the same form as above. It can be shown that this space is also a Frechet space and additionally has the Heine-Borel property, so it follows that it is not normable (the metric is not induced by a norm). On the space $C^\infty(0,1)$ is constructed the space of distributions with compact support, which are linear functionals that are continuous with respect to the above defined topology.
To generalise the above, you should replace $(0,1)$ with an arbitrary open subset of $\Omega\subset\mathbb{R}^m$, and allow $k$ to be a multi-index.
Other continuous function spaces that you might be interested in are:
(i) the space $\mathcal{D}_K$ which is the space of all $f\in C^\infty(\mathbb{R}^m)$ such that $\text{supp}f\subset K$ where $K\subset\Omega$ is compact. It can be shown that $\mathcal{D}_K$ is a closed subspace of $C^\infty(\Omega)$.
(i) the Schwartz space $\mathcal{S}(\mathbb{R}^m)$, which is important for Fourier analysis. On this space is consructed the space of tempered distributions, which is composed of linear functionals that are continuous with respect to the appropriate topology defined on the Schwartz space.
(ii) the space of test functions $\mathcal{D}(\Omega)=C^\infty_0(\Omega)$, which is important for Sobolev space and PDE theory. The space of distributions is constructed on this space, and consists of all linear functionals that are continuous with respect to the appropriate topology on $\mathcal{D}(\Omega)$. This topology is a challenge to define and understand, but is easily characterised in terms of convergence.
By far the best reference for all of this stuff is Rudin's book on functional analysis. Also see Yosida's book on functional analysis, DiBenedetto's book on real analysis and Knapp's book on advanced analysis.
Let $f$ be a continuous function from a metric space $X$ into $Y$. If $V\subset Y$ and $V$ is open, then we shall prove that $f^{-1}(V)$ is open.
Suppose that $p\in X$ and $f(p)\in V$. Since $V$ is open, there exists $\varepsilon>0$ such that $y=f(x) \in V$ if $d_{Y}(f(x),f(p))<\varepsilon$, and since $f$ is continuous at $p$, there exists $\delta>0$ such that $d_{Y}(f(x),f(p))<\varepsilon$ if $d_{X}(x,p)<\delta$. Thus $x\in f^{-1}{(V)}$ as soon as $d_X(x,p)<\delta$.
Best Answer
You have to think in terms of the subspace topology. Let $\tau$ be the usual topology on $\mathbb R$ and $S=[0,1]$ and $A=[0,1/2)$. Then $A = S\cap U$, where $U$ is for example $(-1/2,1/2)$ (which is an element of $\tau$), so $A$ is open in the subspace topology, which is defined as $\tau_S = \{S\cap U : U\in \tau\}$.