Can Infinite Intersections of Open Sets in Metric Spaces Be Closed?

[6.28318531]

Homework Statement

Find (X,d) a metric space, and a countable collection of open sets U\subsetX
for i \in Z^{+} for which
\bigcap^{∞}_{i=1} U_i
is not open

Homework Equations

A set is U subset of X is closed w.r.t X if its complement X\U ={ x\inX, x\notinU}

The Attempt at a Solution

Well I don't really know where to begin, I suppose I could see why an infinite intersection of sets could be closed, but how do I begin?

 

[80past2]

What happens if M = R, and Ui = the open ball of radius 1/i centered at 0?

 

[6.28318531]

Sorry M=R?

So the infinite collection would be open balls all ""stacked up"" inside each other, then they would all intersect, and that intersection would be closed, you could never get outside of the smallest ball, even though they are all open, you can keep on choosing a smaller ball? I can visualise what's happening, but how do you write that, do you use the definition of the open ball, and set d(0,x) <1/i ?

 

[Fredrik]

It's not obvious that the intersection of a decreasing ($B_1\supset B_2\supset\dots$) sequence of open balls is not open. You should think of simple set that you know is not open, and try to express that as a countable intersection of open sets. (The idea to use a decreasing sequence of open balls will work, but it's not your only option).

I use the notation B(x,r) for the open ball around x with radius r. This notation might be useful here.

 

[6.28318531]

Well I can visualise the ball idea, it does make sense in a way... So are you saying try something like [0,1] w.r.t ℝ as a countable intersection of sets

 

[Fredrik]

Yes, that's a good set to use as the starting point. A singleton set like {0} would work too.

 

[6.28318531]

Thanks for the help so far..I will have to get back to this tomorrow.

 

[6.28318531]

So let X =R, d some trivial metric the intersection U = {0}, and each Ui a subset of R as long as 0 element Ui?

 

[Fredrik]

I don't quite understand. What do yo mean by "as long as 0 element Ui"?

Anyway, you don't have to pick some trivial metric. The standard metric will do just fine. (The d defined by d(x,y)=|x-y|). Just find a sequence $\langle E_i\rangle_{i=1}^\infty$ (I suggest a decreasing sequence, $E_1\supset E_2\supset\dots$) of open sets such that $\bigcap_{i=1}^\infty E_i=\{0\}$.

 

[micromass]

Something I see in this thread is that you seem to think that "a set is not open" is the same as "a set is closed".

This is ABSOLUTELY FALSE. A set can be neither open and closed, and a set can be both open and closed. Remember very well that "not open" is not the same as "closed"!

 

[6.28318531]

@Fredrik; Oh yeah I can see what you mean, I wrote that in a hurry, I should have thought a bit more before I typed. What about Ui=[0,1/i]?

@Micromass; I sort of understand how a set be open and closed at the same time, is ∅ an example?

 

[HallsofIvy]

Yes, but more importantly "most" sets are neither open nor closed.

 

[Fredrik]

What about Ui=[0,1/i]?

The intersection of those sets is {0}. But the U_i you chose aren't open. Can you do something similar with open intervals?

 

[6.28318531]

So maybe something like [0] U (0, 1/i) ?

 

[Fredrik]

If you meant {0}U(0,1/i), then this is equal to [0,1/i). This is not an open set. A set is open if and only if all of its points are interior points. 0 is not an interior point of [0,1/i).

 

[6.28318531]

I'm sorry this must be starting to drag out...how about (-1/i,1/i)?

 

[Fredrik]

Don't worry about it. I didn't take me a lot of time to write these replies, and we're at least making progress.

Those are open sets. Can you prove that their intersection is {0}, and that {0} is not open?

 

[6.28318531]

Well if {0} has an open complement it is closed, so the interval (-∞,0) U (0, ∞) which is open, which means that {0} is closed. As for showing that U={0} = \bigcap^{∞}_{i=1} Ui , Ui=(-1/i,1/i), well you want to show that for some x\in\bigcap U, and x\in U1, U2,U3...Ui, the only number that satisfies this is zero.
So for some n→∞ then Un→(0,0) so Un→{0}. Also by observation 0 \in U1,U2,...Un-1, by the definition of Ui

so 0 is the only number that can be in the intersection U1\capU2\cap...\capUn, so the intersection is the single {0}.

 

[Fredrik]

Well if {0} has an open complement it is closed, so the interval (-∞,0) U (0, ∞) which is open, which means that {0} is closed.

This is correct, but not sufficient. I wouldn't call (-∞,0) U (0, ∞) an "interval", but you're right that this set is open. You didn't show us how we can be sure that it's open, but the real problem with your proof is that the result that {0} is closed doesn't immediately imply that {0} isn't open (assuming that you haven't already proved that all closed subsets of ℝ are not open). (There are metric spaces that have subsets that are both open and closed).

As for showing that U={0} = \bigcap^{∞}_{i=1} Ui , Ui=(-1/i,1/i), well you want to show that for some x\in\bigcap U, and x\in U1, U2,U3...Ui, the only number that satisfies this is zero.
So for some n→∞ then Un→(0,0) so Un→{0}. Also by observation 0 \in U1,U2,...Un-1, by the definition of Ui

so 0 is the only number that can be in the intersection U1\capU2\cap...\capUn, so the intersection is the single {0}.

This is not a valid proof. I don't understand some of the things you're saying, in particular "for some n→∞". The standard way to prove an equality between two sets, say A=B, is to first prove that if x is in A, then x is in B, and then prove that if x is in B, then x is in A. These two results imply that A and B have the same members, and that means that A=B.

In this case, I wouldn't be that formal, since the equality is obvious if you understand what $\bigcup_{i=1}^\infty(-1/i,1/i)$ means. (It's obvious that 0 belongs to all the (-1/i,1/i) and that no other real number does). However, if you're not very familiar with the formal way of proving these things, I suggest that you try it.

 

[6.28318531]

Yeah I have only just started a course doing things "properly".What I was trying to say as n grows sufficiently large Ui decreases to {0}. I didn't really know the best way to write this. I was trying to use an argument based on contradiction,


For showing something is not open Do we have to use an open ball? So an argument like; {0} is not open w.r.t R if you consider {0}\subsetℝ. So if you consider the open ball centred at 0, then For every ε> 0 the open Ball B= {x\inℝ, | 0-x| < ε} contains the point -ε/4, which is not in {0}, so there is no such open ball, so {0} is not open.

Thanks

 

[Fredrik]

That's a good way to show that 0 is not an interior point of {0}. If your book defines "open set" as a set E such that every member of E is an interior point of E, or if it proves a theorem that says that a set E is open if and only if every member of E is an interior point, then your proof also shows that {0} is not open (and it's probably the easiest way to do it).

 

[6.28318531]

Okay thanks for that then, you helped alot