Why is the Empty Set Open in a Metric Space?

Click For Summary

Discussion Overview

The discussion revolves around the properties of open sets in metric spaces, specifically addressing why the empty set is considered open. Participants explore the implications of definitions and theorems related to continuity and openness, as well as the nature of boundaries in metric spaces.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the empty set is open in any metric space, questioning whether this is a vacuous truth.
  • Others argue that if the empty set were not open, it would imply the existence of a point within it, leading to a contradiction.
  • Concerns are raised about the continuity of functions when proving sets are open, particularly regarding the function f(x,y)=(1/x)-y, which is not continuous at x=0.
  • Some participants clarify that continuity is only considered on the domain where the function is defined, suggesting that f can be continuous on a restricted domain.
  • There is a discussion about the nature of open balls in metric spaces, with some participants emphasizing that all points in an open ball are contained within the space itself, thus reinforcing that the space is open.
  • Participants explore the implications of the definition of open sets, particularly how the empty set satisfies the conditions for being open due to the false antecedent in the logical implication.
  • Another argument presented states that since the entire space X contains all its limit points, it is closed, making its complement (the empty set) open.

Areas of Agreement / Disagreement

Participants generally agree that the empty set is open, but there is ongoing debate regarding the implications of this property and the conditions under which functions are considered continuous. The discussion remains unresolved regarding the nuances of continuity and the application of theorems in specific contexts.

Contextual Notes

Participants express uncertainty about the assumptions required for the continuity of functions and the implications of definitions of open sets. There are unresolved questions about how to appropriately apply theorems in the context of discontinuities.

kingwinner
Messages
1,266
Reaction score
0
1) Fact: Let X be a metric space. Then the set X is open in X.
Also, the empty set is open in X.

Why??



2) Let E={(x,y): x>0 and 0<y<1/x}.
By writing E as a intersection of sets, and using the following theorem, prove that E is open.
Theorem: Let X,Y be metric spaces. If f:X->Y is continuous on X, then f -1(V) is open in X whenever V is open in Y.
Proof:
E = {(x,y): x>0} ∩ {(x,y): y>0} ∩ {(x,y): (1/x)-y>0}
A finite intersection of open sets is open, so it's enough to show each of the 3 sets are open.
To prove that {(x,y): (1/x)-y>0} is open, let f(x,y)=(1/x)-y.
Then {(x,y): (1/x)-y>0}=f -1(0,∞) is open because the interval (0,∞) is open and f is continuous.
...

Now I don't understand why we can say that f is continuous and use the above theorem. When we break E into 3 sets, each set is to be treated separately and independently, so for {(x,y): (1/x)-y>0}, we don't assume x>0, but then f will not be continuous (f is not continuous at x=0), then how can we use the above theorem to prove that {(x,y): (1/x)-y>0} is open?? I don't understand...

Can someone please explain?
Thanks for any help!
 
Physics news on Phys.org
1)Consider an open ball in X. Are all of the points of the open ball in X? If the empty set was not open, then there would have to be a point in the empty set satisfying some condition (you can verify the negation yourself), which is contradictory.

2) Use a little common sense. Of course f is not continuous where it isn't defined. Whenever we say a function is continuous, we mean continuous on its domain of definition.
 
snipez90 said:
1)Consider an open ball in X. Are all of the points of the open ball in X? If the empty set was not open, then there would have to be a point in the empty set satisfying some condition (you can verify the negation yourself), which is contradictory.

2) Use a little common sense. Of course f is not continuous where it isn't defined. Whenever we say a function is continuous, we mean continuous on its domain of definition.
1) "X is open in X."
But when you get to the boundary, and put an open ball around it, wouldn't there be some points outside of X?

"the empty set is open in X"
But the empty set doesn't even have any elements. Is this statement like a vacuous truth kind of thing?


2) Consider the set {(x,y): f(x,y)=(1/x)-y>0} = f -1(0,∞) . We want to prove that it is open by using the theorem.
My point is, does it even SATISFY the conditions of the theorem? The theorem requires f to be continuous everywhere, but f(x,y)=(1/x)-y is certainly not continuous everywhere. And indeed, x=0 is a discontinuity (it's an infinite discontinuity). Looking at the set {(x,y): f(x,y)=(1/x)-y>0} separately on its own, it does NOT include the additional assumption that x>0.
So I think not all of the assumptions in the theorem are satisfied...how can we fix this problem?

Thanks for any help!
 
Last edited:
1) "X is open in X."
But when you get to the boundary, and put an open ball around it, wouldn't there be some points outside of X?

There is no boundary of the whole space! There is nothing outside of the whole space.
 
kingwinner said:
"the empty set is open in X"
But the empty set doesn't even have any elements. Is this statement like a vacuous truth kind of thing?

There may be some fine points I'm not aware of, but I would say yes its mostly to make other mathematical theorems look nicer/simpler. Less explicit mention of the empty set as a special case.

Torquil
 
rochfor1 said:
There is no boundary of the whole space! There is nothing outside of the whole space.

So I guess the key point is that balls of radius r about a are defined as {x E X: d(x,a)<r}, so that no matter how big r is, any ball is ALWAYS contained in X, so X is always open in X.
 
rochfor1 said:
There is no boundary of the whole space! There is nothing outside of the whole space.

So I guess the key point is that balls of radius r about a are defined as {x E X: d(x,a)<r}, so that no matter how big r is, any ball is ALWAYS contained in X, so X is always open in X.
 
kingwinner said:
2) Consider the set {(x,y): f(x,y)=(1/x)-y>0} = f -1(0,∞) . We want to prove that it is open by using the theorem.
My point is, does it even SATISFY the conditions of the theorem? The theorem requires f to be continuous everywhere, but f(x,y)=(1/x)-y is certainly not continuous everywhere. And indeed, x=0 is a discontinuity (it's an infinite discontinuity). Looking at the set {(x,y): f(x,y)=(1/x)-y>0} separately on its own, it does NOT include the additional assumption that x>0.
So I think not all of the assumptions in the theorem are satisfied...how can we fix this problem?

So now I'm only left puzzled with this one...

Could someone explain the deatils of this?
Thanks for any help!
 
kingwinner said:
So now I'm only left puzzled with this one...

Could someone explain the deatils of this?
Thanks for any help!

f is defined and continuous when restricted to S = {(x, y) | x is not 0} , which is all we care about.

In more detail:

You can easily show that S is open. Consider the metric space (S, d'), where d' is the metric of X restricted to S. You have f : S -> Y, and f is continuous on S. So f^-1 (0, ∞) is open in S. Since S is open in X, f^-1 (0, ∞) is open in X.
 
  • #10
kingwinner said:
So I guess the key point is that balls of radius r about a are defined as {x E X: d(x,a)<r}, so that no matter how big r is, any ball is ALWAYS contained in X, so X is always open in X.

It sounds like you fully understand why X is open in X. Just to maybe add why the empty set is open in X, consider the following:
empty set open in X iff for every x in the empty set there exists r>0 s.t B(x,r) contained in
empty set
Obviously the antecedent of the above implication is false so any consequent will follow (asides from those contradicting the antecedent) making the implication true (empty set open in X)
eg: the moon is made of green cheese then Hitler is alive
is a true implication since the antecendent (moon made of green cheese) is always false.
For a mathematical proof, consider modus tollens of the implication.
And if you're really bored, you can prove modus tollens is equivalent to the implication via proof by contradiction.
 
  • #11
Another argument justifying why the empty set is open: X contains all of its limit points, so X is closed. Therefore its complement, the empty set, is open.
 

Similar threads

Undergrad On inverse function theorem in Spivak's CoM
  • · Replies 6 ·
Replies
6
Views
4K
Undergrad Find f(z) given f(x, y) = u(x, y) + iv(x,y)
  • · Replies 8 ·
Replies
8
Views
1K
Undergrad Determination of error in interpolating polynomial
  • · Replies 5 ·
Replies
5
Views
3K
Undergrad What is the difference between these two power series theorems?
  • · Replies 5 ·
Replies
5
Views
3K
Undergrad Proving that convexity implies second order derivative being positive
  • · Replies 6 ·
Replies
6
Views
3K
Undergrad Why prove Taylor's theorem with p(x)=q(x)−((b−a)^n/(b−x)^n)q(a)?
  • · Replies 9 ·
Replies
9
Views
3K
High School Why is this definite integral a single number?
  • · Replies 20 ·
Replies
20
Views
4K
Undergrad Finding the Largest (or smallest) Value of a Function - given some constant symmetric errors
  • · Replies 3 ·
Replies
3
Views
2K
MHB Implicit function theorem for f(x,y) = x^2+y^2-1
  • · Replies 1 ·
Replies
1
Views
2K
High School Calculating shaded area: I'm getting discrepancies between methods
  • · Replies 3 ·
Replies
3
Views
3K