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's important to understand that when talking about topology open set means a set what belongs to the topology. You should't think about it as something 'open' in any other sense.
So of course the most common example is:
1) All open (in traditional sense) subsets of $R^n$ form a topology over $R^n$
There are trivial examples like:
2) For any set, all subsets form a topology. Then by definition all subsets are both open and closed.
There are also harder to graps topologies, like:
3) Zariski topology is the topology for which closed sets are all subsets of, which are zeroes of some polynomial.
For the real line this means that all finite sets of points are closed. This is also called "finite complement topology".
4) On the real line there is another topology called "lower limit topology". It is defined such that open sets are all half open intervals $[a, b)$ (and therefore all their unions).
It is a useful exercise to prove that all those examples are indeed topologies, i.e. they satisfy the definitions.
Best Answer
There's a misunderstanding: neither compactness nor openness is a property of a set. In fact given a set $X$ we define what subsets are open (they have to obey topology axioms) on that set. If you see someone stating that, say "$(0,1)$ is open in $\mathbb{R}$" then he is implicitly assuming the Euclidean topology on $\mathbb{R}$. You always start with topology (meaning the collection of all open subsets on a given set).
Now compactness is a property of a particular topology. So you first start with open sets and then you can conclude whether the space is compact or not. If we change open subset then other properties (compactness, connectedness, etc.) may change as well.
Also we can do whatever we want in maths. We defined what "open sets", or more precisely "topology", means. And then given a set $X$ we can consider different topologies on that set. We can create them however we want, there's nothing that can stop us. And then we can analyze properties of such spaces like compactness.
For example $\mathbb{R}$ can be given multiple different topologies:
and so on, and so on, in fact infinitely many. We then ask what properties each of these spaces has? Is it compact? Connected? Etc. For example 1. is connected but not compact, 2. is neither compact not connected, 3. is compact and connected, and so is 4.
For a given (logically valid) combination of topological properties, you will probably find a topology on $\mathbb{R}$ which satisfies them.
In some cases yes. For example consider the set of all real numbers $\mathbb{R}$. Typically when topology is not explicitly defined then the Euclidean topology is assumed.
Analogously $\mathbb{R}^n$ (and its subsets), is typically considered with the product topology (respectively subspace topology) based on the Euclidean $\mathbb{R}$. Unless explicitly stated otherwise.
But when dealing with non standard sets, some general $X$, then it is expected from the author to clearly state what topology he is considering. Otherwise it may lead to confusion.
That's the other way around. We don't start with calculus and then define open subsets. We start with open subsets and slowly build entire calculus machinery. For example convergence is meaningless without open subsets (I mean it can be defined via metric/norm but these are equivalent). And convergence is necessary for derivative definition. Not only that, but we also need subtraction and division, and it would be good if those were to behave nicely (i.e. continuous). Other properties of the Euclidean $\mathbb{R}$ are also of great importance for the calculus, e.g. local compactness, completeness of the Euclidean metric, path connectedness, etc.
On the other hand general topology is not enough to perform differential calculus. Differential calculus is based on $\mathbb{R}$ with the standard Euclidean topology. It can be performed on other spaces ($\mathbb{R}^n$, manifolds, Banach spaces, etc.) but all of those are heavily related to the standard $\mathbb{R}$. There's simply something very special about $\mathbb{R}$ (with the Euclidean topology) making it a very rich and useful structure.
Of course this is not 100% accurate. Some abstract forms of calculus (e.g. derivations) can be done without $\mathbb{R}$ and on other nontopological (e.g. purely algebraic) structures, but lets just assume that these are not mainstream maths. There are also other approaches, e.g. p-adic analysis, see this post: Can we define a derivative on the $p$-adic numbers?
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