Why is the Empty Set Open in a Metric Space?

[kingwinner]

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!

 

[snipez90]

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.

 

[kingwinner]

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!

 

[rochfor1]

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.

 

[torquil]

"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

 

[kingwinner]

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]

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]

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!

 

[Werg22]

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.

 

[bassbud101]

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.

 

[jbunniii]

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.