Too long for the comments:
You are confusing a circle with a disc. A circle in a topological context is (homeomorphic to) the boundary of a disc. A circle has empty interior, in part because it is a boundary. One can of course speak about the region of $\Bbb R^2$ that is enclosed by the circle as, in some sense, its "interior". However, when you embed a circle in say $\Bbb R^3$, this notion of interior no longer makes sense, while $S^1 \subseteq \Bbb R^3$ still has empty (topological) interior.
Furthermore, neither a circle, nor a disc is open and closed in the standard topology on $\Bbb R^2$. The circle is closed, as is the disc. You should try and convince yourself that they are not open.
There is no problem with using a metric to define a topology. And you don't need a topology to define a metric space, you just need a metric. Surely you don't need to talk about open sets to construct a function $d: X \times X \to \Bbb R$?
A topological space is not really "more fundamental," it just happens that every metric space has a topology that is compatible with the metric in some sense. This is what we should want to happen, since topological spaces are generalizations of metric spaces.
You should also bear in mind that the definition of what an open set is depends on what sets you choose to be open. There are many different ways of choosing those open sets, many of which are compatible with various other mathematical structures. In addition to metric spaces, if a set has a linear order, then we can define in a particular way, a topology on that set using the linear order. When we look at algebraic objects, like groups, it might be possible to find a topology on the underlying set such that the group operation the map $g \mapsto g^{-1}$ are continuous. (This is called a topological group.)
In fact, $\Bbb R$ with the standard topology is an example of all of these. It is a metric space, linearly ordered, and is a group under addition. Just to be clear, the standard topology on $\Bbb R$ is generated by the Euclidean metric, but can also be generated by the linear order $<$, and makes addition continuous. This single, easily defined topology allows us to use three (actually more) extremely useful objects when discussing the topology of $\Bbb R$.
Remember that a topology on a point-set $X$ is just a collection of subsets of $X$ such that the collection satisfies certain properties. Look at the abstract definition of a topology. It only requires three axioms:
$\emptyset$ and $X$ are in the topology.
That the collection is closed under arbitrarily large unions.
That the collection is closed under finite intersections.
So that it is easier to talk about members of the collection, we call them open sets.
Observe that nowhere in the definition of what a topology is do we give any characterizations about the open sets themselves. This is somewhat analogous to how the axioms of a group make no mentions of what the group elements are, only how they interact.
The advantage to this flexibility is that it is easy to put different kinds of topologies on an arbitrary point-set. The down-side is that often we have too many choices and the easy ones (the discrete and trivial topologies) are rarely useful or what we want (but sometimes they are!).
Let's say you have a point-set $X$ and you want to put a topology on it. Chances are that you have some additional structure on $X$ and you want to use topological tools to study it. In that case, whatever topology you pick needs to be compatible with that structure in some sense that will vary a lot between structures.
The prototypical example is the metric topology. Suppose $(X,d)$ is a metric space. From analysis we have a definition of continuity on $X$, but we also have a different definition for arbitrary topological spaces:
Analytic Continuity: A function $f: X \to Y$ is continuous at $x \in X$ if for all $\epsilon > 0$, there exists a $\delta > 0$ such that for all $y \in X$, if $d_X(x,y) < \delta$, then $d_Y(f(x),f(y)) < \epsilon$.
Topological Continuity: A function $f: X \to Y$ is continuous if for all open sets $U \subseteq Y$, the preimage $f^{-1}(U)$ is open in $X$.
Whatever choice we make of topology on $X$, we want these two definitions to be equivalent, so that we can combine the analytic and topological machinery.
The take-away of all this is that when you just have a point-set and nothing else, it's not clear what the open sets should be or why you would even need these open set doohickeys. In fact, when we make purely abstract point-set topology arguments, it doesn't matter what an open set is, since we are reasoning about them abstractly. Once we start looking at (slightly) more concrete mathematical structures, like say metric spaces or groups, we can no longer reason about open sets abstractly, since there are certain compatibility properties we would like them to satisfy. Now we need to single out a particular collection of subsets that we would like to call the open sets and show that is forms a topology.
The other sensible choice of what a continuous function is what we call an open function:
- Open Function: A function $f: X \to Y$ is called open if for all open sets $U \subseteq X$, the image $f(U)$ is open in $Y$.
So why don't we call these functions continuous?
The short (unhelpful) answer is because the preimage definition is what works. To give a precise mathematical reason is not easy, because essentially our choice was based on what got us what we wanted. Any good definition of continuity ought to formalize our intuitive notions of continuity and the preimage definition does that better than the image definition.
It is also worth pointing out that preimages play much nicer with set operations than do images. To wit:
$f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$, whereas $f(A \cap B) \subseteq f(A) \cap f(B)$ and
$f^{-1}(A \setminus B) = f^{-1}(A) \setminus f^{-1}(B)$, whereas $f(A \setminus B) \supseteq f(A) \setminus f(B)$.
And since whichever one choose in the definition will be used a lot with sets and set operations, so from a practical standpoint, we kind of want the preimage to be the right definition.
It has become little bit lengthy because I tried to avoid the use of sequences of functions.
We apply the general definition of a limit point:
"$f$ is a limit point of the set $F$ iff for all $\epsilon>0$ there exists a $g\in F$ with $f\neq g$ such that $\Vert f-g\Vert <\epsilon$".
Now we take an arbitrary limit point $f$ of $F$ and show that $f<0$ is impossible.
Let's assume $\epsilon>0$ and $f(z)=-\epsilon$ for an arbitrary $z\in[0,1]$. As stated above, we know that there exists a $g\in F$ with $g\neq f$ and $\Vert f-g\Vert = \sup\limits_{x\in[0,1]}|f(x)-g(x)| <\epsilon$. Note that $g(x)\geq 0$ for all $x\in[0,1]$. This leads to a contradiction because then we would have $\epsilon<|-\epsilon-g(z)|=|f(z)-g(z)|\leq\sup\limits_{x\in[0,1]}|f(x)-g(x)| <\epsilon$. So it must hold that $f(x)\geq 0$ for all $x\in[0,1]$ as $z$ was arbitrarily chosen.
Now we check that $f$ is continuous:
We know that every $g\in F$ is continuous on a closed intervall and hence uniformous continuous. So we know that for a given $\epsilon>0$ there exists a $\delta>0$ such that for all $x,y\in[0,1]$ with $|x-y|<\delta$ it holds $\vert g(x)-g(y)\vert<\epsilon$.
Let's choose an arbitrary $\epsilon>0$ and define $\epsilon':=\frac{\epsilon}{3}>0$. So there exists a $g\in F$ with $\Vert g-f\Vert <\epsilon'$. Moreover we know that there exists a $\delta>0$ such that for every $x,y\in[0,1]$ with $|x-y|<\delta$ it holds $\vert g(x)-g(y)\vert<\epsilon'$ (uniformous continuity of $g$). Let be $a\in[0,1]$ arbitrary but fixed and $x\in[0,1]$ with $|x-a|<\delta$. Then we have $$\vert f(x)-f(a)\vert \leq \vert f(x)-g(x)\vert +\vert g(x)-g(a)\vert+ \vert f(a)-g(a)\vert \leq \\ \Vert f-g\Vert +\vert g(x)-g(a)\vert+ \Vert f-g\Vert<\epsilon'+\epsilon'+\epsilon' = \epsilon.$$
So $f$ is continuous on $[0,1]$ because for an abritrary $\epsilon>0$ we have found a $\delta>0$ such that for all $x\in[0,1]$ with $|x-a|<\delta$ it holds that $|f(x)-f(a)|<\epsilon$.
So we can conclude that $f\in F$. As $f$ was arbitrarily chosen $F$ is closed.
Best Answer
Let $f_n$ be the function whose graph is given by joining the points $(0,0), ({1 \over n}, 1), ({2 \over n}, 0), (1,0)$. Let $f= 0$.
Then $f_n \to f$ with distance $p$, but $d(f_n,f) \to 1$.
Also, if we let $Lf = \int_0^1 f(x)dx$ then we have $|Lf-Lg| \le p(f,g) \le e \cdot d(f,g)$ and so $L$ is continuous with respect to both metrics. Hence $A=L^{-1}((1,\infty))$ is open.