I thought about it for a long time, but I can't come up some good ideas. I think that empty set has no elements,how to use the definition of an open set to prove the proposition. The definition of an open set: a set S in n-dimensional space is called open if all its points are interior points.
$\begingroup$
$\endgroup$
8
-
1$\begingroup$ From what point of view? Topology? $\endgroup$– eccstartupCommented Jul 19, 2013 at 6:10
-
37$\begingroup$ @eccstartup What other point of view is there? $\endgroup$– InkCommented Jul 19, 2013 at 6:14
-
4$\begingroup$ The answers to Reason for “all” and “any” result on empty lists contain some good examples of vacuous truths. $\endgroup$– detlyCommented Jul 19, 2013 at 6:45
-
17$\begingroup$ Do not forget it's also a closed set :) $\endgroup$– chxCommented Jul 19, 2013 at 7:14
-
1$\begingroup$ Much ado about nothing $\endgroup$– Ryszard SzwarcCommented Mar 22, 2022 at 21:51
|
Show 3 more comments
12 Answers
$\begingroup$
$\endgroup$
10
It's vacuously true. All points $\in\emptyset$ are interior points, have a blue-eyed pet unicorn, and live in Surrey.
answered Jul 19, 2013 at 6:10
-
3
-
34$\begingroup$ Strangely enough, they are blue-eyed unicorns themselves! $\endgroup$ Commented Jul 19, 2013 at 10:03
-
2$\begingroup$ I'd be interested to hear a response to @Jellyfish's point (in his answer below): with this argument, you can claim the empty set is closed (it contains all it's limit points, even the red-eyed unicorns ;). $\endgroup$– drevickoCommented Jul 19, 2013 at 10:56
-
11$\begingroup$ @drevicko: There is nothing wrong. Given a topology over a set, both the empty set, and the set itself, are both closed and open. Remember that a set can be neither closed, nor open (e.g. the [0, 1[ interval in R in its natural topology) $\endgroup$ Commented Jul 19, 2013 at 11:39
-
1$\begingroup$ @HagenvonEitzen Hm, that begs for another discussion if we can visualize some concepts in matematics. The notion of "seven-dimensioned sphere with 5 handles" still boggles my mind. $\endgroup$ Commented Jul 20, 2013 at 13:20
$\begingroup$
$\endgroup$
5
Here's another perspective:
Suppose the empty set is not considered open.
Consider the sine function $\sin\colon \Bbb R \to \Bbb R$.
Then $(3\,.\,.\,4)$ is open (presumably), but $\sin^{-1}(3\,.\,.\,4)=\varnothing$ is not open, so the sine function is not continuous.
We don't want that!
Another thing: one of the key properties of open sets is that the intersection of any two open sets is open. If the empty set were not considered open, then that wouldn't be true anymore.
-
6$\begingroup$ Actually, this is not a proof. $\endgroup$ Commented Jul 19, 2013 at 12:59
-
39$\begingroup$ It's not supposed to be a proof, per se. It's giving a couple reasons for the definition of a topology to require the empty set to be open. $\endgroup$– dfeuerCommented Jul 19, 2013 at 15:59
-
1$\begingroup$ I proved that it's not a requirement; the only axioms are closure for unions and finite intersections (and this gives you a topology for the union over the family of open sets). $\endgroup$ Commented Jul 19, 2013 at 18:15
-
1$\begingroup$ @EdoardoLanari Yes, so you did. I do in fact find your argument compelling. But your argument is not a proof either. BOTH you and dfeuer have answers addressing the question "what is the best set of axioms" and not "what follows from which axioms". $\endgroup$ Commented Oct 5, 2015 at 18:42
-
$\begingroup$ In order for something to be a proof of something else you obviously need to putb things into a framework. I assume Kelley's definition of a topology on a set and I prove the empty set to be open. Assuming the more familiar axioms it is true by definition, so I don't get your comment. Or, better, I fear it is due to thinking this situation from an analyst point of view, though I might be wrong. $\endgroup$ Commented Oct 5, 2015 at 19:33
$\begingroup$
$\endgroup$
2
My answer would have nothing to compare with @HagenvonEitzen's wit and to the point, but here's a way to think of it:
The complement of an empty set is the whole set, which of course contains everything including all limit points. Hence the whole set is closed, and therefore it compliment, empty set is open.
Hagen's argument can be made to show empty set is closed and whole set is open. Because all points in empty sets are limit points, so empty set is closed. So its compliment, whole set is open.
Another (better) way to think of this:
A topological space generalizes the concept of a metric space. With this view, a function is continuous iff the inverse image of an open set is open. Since a function that maps the entire space onto a single point is always continuous, the empty set is open. Take an open set which does not contain the single point. Its inverse image is the empty set.
Above is a proof for the definition, however, empty set is open by the definition of a topology. But it's good to know why things are defined the way they are.
A final note: It should be "the" empty set, since it is unique.
answered Jul 19, 2013 at 6:38
-
$\begingroup$ oh, you use the knowledge of closed sets to prove it.Considering the complement of an closed set is an open set, I think your answer is right. $\endgroup$– python3Commented Jul 19, 2013 at 6:49
-
$\begingroup$ Yes, I used the compliment. Actually I think this was my homework question, and this is approximately how I answered. =) $\endgroup$ Commented Jul 19, 2013 at 6:52
$\begingroup$
$\endgroup$
4
Indeed, all the points of the empty set are interior points.
This is a statement that is ''vacuously true".
In other words: If you produce an element of the empty set, I will be happy to show that it is an interior point. This can be done because first of all you can't produce one. Compare to Russel's assertion in an apocryphal tale, "Given an inconsistent proposition, I can prove any statement"(paraphrased).
-
3
-
-
$\begingroup$ ok, I know this is probably beating a dead horse but I have to ask. Why is it considered vacuously true? I usually think of vacuously true in terms of the antecedent being false, but there is not implication in the definition. So how does "vacuously true" apply here? $\endgroup$ Commented Jun 20, 2018 at 2:52
-
$\begingroup$ @CharlieParker Here is OP's definition of "$U$ is open" expressed as an implication: "For all points $x$, if $x\in U$ then $x$ is an interior point of $U$." If $U=\emptyset$ then, for any point $x$, the antecedent $x\in U$ is false. (Just in case the horse is still alive two years later.) $\endgroup$ Commented Jul 24, 2020 at 13:04
$\begingroup$
$\endgroup$
2
Another way to think about it is that, since the empty set has no points, you can't find a point in the empty set that is NOT an interior point to violate the definition. So it is true "by default," or vacuously.
-
$\begingroup$ ok, I know this is probably beating a dead horse but I have to ask. Why is it considered vacuously true? I usually think of vacuously true in terms of the antecedent being false, but there is not implication in the definition. So how does "vacuously true" apply here? $\endgroup$ Commented Jun 20, 2018 at 2:53
-
$\begingroup$ The statement is vacuously true for the empty set because the empty set does not contain any point (hence vacuous like vacuum), and therefore the empty set is open because the nonexistence of points in it guarantees that it can't possibly violate the definition for an open set that all its points are interior points. Put another way, the condition that all points are interior points is satisfied NOT by every point being an interior point. The condition is satisfied vacuously by no points existing in the set in the first place to not be an interior point. $\endgroup$– Ming HoCommented Jul 13, 2018 at 22:40
$\begingroup$
$\endgroup$
2
It's the union over the empty subfamily of the topology: that's all.
-
1$\begingroup$ John L. Kelley's General Topology begins “A topology is a family $\def\I{\mathfrak J}\I$ of sets which satisfies two conditions: the intersection of any two members of $\I$ is a member of $\I$, and the union of the members of each subfamily of $\I$ is a member of $\I$.” He doesn't mention the empty set specifically, allowing it to be implied by the second condition, as you said. (He also doesn't mention the underlying space; it too is implicit, as $\bigcup_{U\in\I} U$.) $\endgroup$– MJDCommented May 30, 2014 at 4:55
-
$\begingroup$ I had in mind that book while I was writing my answer!! $\endgroup$ Commented May 30, 2014 at 6:15
$\begingroup$
$\endgroup$
1
A set is open or closed (or neither) inside another set (actually a set, equipped with a topology).
By the definition of topology (https://en.wikipedia.org/wiki/Topology#Mathematical_definition) the empty set is always open and its complement (i.e. the whole other set) is always closed.
Therefore is the emtpy set open and closed in every topology.
answered Jul 19, 2013 at 12:15
-
1$\begingroup$ It may be a little late, but despite its low up-votes I actually think this is the most correct answer. To talk about being open, one must have ascribed a topology to the space. Having done that, the empty set is always DEFINED to be open. This discussion of interior points and vacuous truths can only be done after the fact; that is, only after you know what an open set is! $\endgroup$ Commented Jan 9, 2014 at 19:37
$\begingroup$
$\endgroup$
A different way to think of why this absolutely has to be true in spaces with some disjoint open sets (which is definitely the case in $\mathbb R^n$) is that the intersection of any finite collection of then must be open. Clearly in these spaces the empty set arises in this way. This is motivation, not a complete argument. The fact that $\emptyset=\bigcup_{A\in\emptyset}A$ is a better reason.
Note that these are general topological answers, not really tied to real spaces.
answered Jul 20, 2013 at 5:26
$\begingroup$
$\endgroup$
5
Because it is the compliment of a closed set.
answered Oct 4, 2013 at 18:14
-
-
$\begingroup$ @JLA: A set that contains all its limit points is a close set (like the complement of $\emptyset$ that contains all points) $\endgroup$ Commented Oct 15, 2013 at 3:36
-
1$\begingroup$ Fine, but in general a set is closed if its compliment is open, so your answer is circular. $\endgroup$– JLACommented Oct 15, 2013 at 3:58
-
$\begingroup$ @JLA: Yes, surprisingly enough if you change my definition of a closed set to a definition that makes it circular it actually becomes circular. The complement of $\emptyset$ contains its limit points and contains only interior points and therefore is open and closed. So it makes sense to consider $\emptyset$ as open and closed too though it would be possible to define open and closedness of sets and exclude the emptyset of being closed and open. $\endgroup$ Commented Oct 15, 2013 at 5:40
-
1$\begingroup$ My point was "the" definition of a closed set is one whose complement is open, so by you saying the empty set is open because its compliment is closed raises the question as to what is a closed set. A set is not defined to be open if its compliment is closed, so I still don't think your definition really applies. $\endgroup$– JLACommented Oct 15, 2013 at 5:46
$\begingroup$
$\endgroup$
The way I wrapped my head around it, is by defining a subset $E$ of a metric space $M$ to be open if $\forall x: x\in E \rightarrow x\text{ is an interior point of }E$. If $E$ denotes the empty set $\emptyset$, no $x$ is contained in $E$, so the implication is automatically true for all $x$.
$\begingroup$
$\endgroup$
The empty set is open, by definition. Nothing more to it.
$\begingroup$
$\endgroup$
Consider the following definition of an open set: For every element in A, there exists a ball of the element contained in A. What is the negation of that statement? Well, it is: There exists an element in A such that every ball is not contained in A. Clearly, the empty set does not satisfy the negation of an open set, therefore it must be vacuously true that the empty set satisfies the definition of an open set.