Is empty set open?bounded?perfect?compact?

[kntsy]

open? A set is open if every point is interior point but empty set has no point! So no?

bounded?no point so cannot measure distance?so no?

perfect? i guess no because there is no point in empty set.

compact?There is no open cover "covering" the empty set right? so no?

 

[Hurkyl]

open? A set is open if every point is interior point but empty set has no point! So no?

If there aren't any of them, then obviously all of them are interior points!

If you have trouble with universal quantification over the empty set, you will probably have no trouble understanding the negation.

A set is not open (relative to a given topology) if and only if there exists a point in it that is not an interior point.

 

[Landau]

compact?There is no open cover "covering" the empty set right?

There is, namely the cover consisting of the empty set (but for this you need that the empty set is open, which Hurkyl just explained.) More generally, every finite space is compact (and even more generally, every space with finitely many open sets is compact).

 

[kntsy]

There is, namely the cover consisting of the empty set (but for this you need that the empty set is open, which Hurkyl just explained.) More generally, every finite space is compact (and even more generally, every space with finitely many open sets is compact).

Thanks for the reply? How about bounded? Distance cannot be measured for empty set!

 

[Landau]

Distance cannot be measured for empty set!

Write out precisely the definition of a set being bounded.

 

[kntsy]

Write out precisely the definition of a set being bounded.

E is bounded if there is a real number M and a point q\inX such that d(p,q)<M for all p\inE

But there is no point in empty set so how can d(p,q) be defined?

 

[Landau]

The metric d is a function X\times X\to\mathbb{R}. If X is empty, this is the (unique) empty function.

 

[kntsy]

The metric d is a function X\times X\to\mathbb{R}. If X is empty, this is the (unique) empty function.

Empty function? What is its image? I guess it can be any number?
However, how can we know whether the image of the empty function is smaller than M? The image can be assigned whatever number so can be very big even infinity?
Thanks.

 

[Landau]

The image is empty. Formally, since a function is a relation, you can view it as the the empty relation
\emptyset\subseteq (X\times X)\times\mathbb{R}.
Now X is bounded if there exist M>0 and q\in X such that, for all p\in X, d(p,q)&lt;M. But there is no such q!

 

[kntsy]

The image is empty. Formally, since a function is a relation, you can view it as the the empty relation
\emptyset\subseteq (X\times X)\times\mathbb{R}.
Now X is bounded if there exist M>0 and q\in X such that, for all p\in X, d(p,q)&lt;M. But there is no such q!

I start seeing the picture now but i still do not understand why "empty<M" is wrong. Empty means nothing so "nothing" is smaller than "something"?
I also wonder if the empty set is unbounded because "there does not exist q"?!??

 

[Landau]

I agree it's a bit tricky. You could also say that X is bounded if there exist M>0 such that for all p,q\in X, d(p,q)&lt;M. With this definition the empty set is bounded.

Or put another way, we can regard X, the empty set, as a subset of any (non-empty) metric space, say \mathbb{R}. Then X is bounded if it is contained in some ball of radius>0. But certainly X is contained in, say, the ball (-1,1) of radius 1 centered at 0.

 

[kntsy]

I agree it's a bit tricky. You could also say that X is bounded if there exist M>0 such that for all p,q\in X, d(p,q)&lt;M. With this definition the empty set is bounded.

Or put another way, we can regard X, the empty set, as a subset of any (non-empty) metric space, say \mathbb{R}. Then X is bounded if it is contained in some ball of radius>0. But certainly X is contained in, say, the ball (-1,1) of radius 1 centered at 0.

So a set can be bounded and unbounded at the same time? However, we know that empty set is compact. Doesn't it imply that the implication compact->closed and bounded does not work?

Thanks Landau.

 

[HallsofIvy]

So a set can be bounded and unbounded at the same time? However, we know that empty set is compact. Doesn't it imply that the implication compact->closed and bounded does not work?

Thanks Landau.

No, no one here has said that the empty set is unbounded.

A set, A, in a metric space, is bounded if there exist a number, M> 0 such that "if x and y are in A, then d(x,y)< M". If A is empty, take M to be any positive number at all then the statement "if x and y are in A, then d(x,y)< M" is TRUE because it is an "if then" statement in which the hypothesis "if x and y are in A" if FALSE.

p=> q is true when p is false, whether q is true or not.