Open set challenging question and mind-blogging concept!Welcome.

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"Open" set challenging question and mind-blogging concept!Welcome.

This statement is correct.
[tex]\left\{\left[1,3-\frac{1}{n}\right)\right\}_{n\in\mathbb Z} \text {is an "OPEN" cover of}\left[1,3\right)[/tex]

BUT WHY OPEN?
There are 3 options.

1.All [itex]\left[1,3-\frac{1}{n}\right)[/itex] is the open set in [itex]\mathbb Z[/itex];
or
2.All [itex]\left[1,3-\frac{1}{n}\right)[/itex] is the open set in [itex]\left[1,3\right)[/itex].
or
3.All [itex]\left[1,3-\frac{1}{n}\right)[/itex] is the open set in [itex]\mathbb R^{1}[/itex]

Which one?
 
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Answers and Replies

  • #2
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This should not be mind-boggling. Look up the notion of a relative topology. But you shouldn't even need that to get the answer(s), as intuition should eliminate the incorrect answer(s).
 
  • #3
HallsofIvy
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This statement is correct.
[tex]\left\{\left[1,3-\frac{1}{n}\right)\right\}_{n\in\mathbb Z} \text {is an "OPEN" cover of}\left[1,3\right)[/tex]

BUT WHY OPEN?
There are 3 options.

1.All [itex]\left[1,3-\frac{1}{n}\right)[/itex] is the open set in [itex]\mathbb Z[/itex];
Ridiculous! [itex]\left[1, 3- \frac{1}{n}\right)[/itex] is not even a subset of Z!

or
2.All [itex]\left[1,3-\frac{1}{n}\right)[/itex] is the open set in [itex]\left[1,3\right)[/itex].
Well, [1, 3) is the set you are talking about isn't it?

or
3.All [itex]\left[1,3-\frac{1}{n}\right)[/itex] is the open set in [itex]R^{1}[/itex]
Again, ridiculous! [1, 3- 1/n) is NOT an open set in that topology! The point "1" is a boundary point of [1, 3-1/n) in R because any neighborhood of 1, no matter how small, will contain some points of larger than 1 and so in [1, 3- 1/n), and some points less than 1 and so not in that set.

In (2), above, 1 is NOT a boundary point because there are NO numbers "less than 1" in the space [1, 3).

Which one?
Not so "mind boggling" if you know the basic definitions.
 
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  • #4
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Ridiculous! [itex]\left[1, 3- \frac{1}{n}\right)[/itex] is not even a subset of Z!
Thanks.
This is very challenging as the statement is made very ambigous by clever mathematicans.
I know [2] is always true but is it the REASON?
For [1], open set in sth need to be a subset of sth? I think it is valid to say R is an open set in Q?
 
  • #5
Office_Shredder
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In fact the definitions are made to be very precise by clever mathematicians.

How can R be an open set in Q if R isn't even in Q? It sounds like you need to go back and check what the definitions of these words mean
 
  • #6
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In fact the definitions are made to be very precise by clever mathematicians.

How can R be an open set in Q if R isn't even in Q? It sounds like you need to go back and check what the definitions of these words mean
I am just proposing. But what is the answer then to this brainteaser then?
 
  • #7
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I am just proposing. But what is the answer then to this brainteaser then?
Two possibilities have been eliminated. It's not hard to find the answer, and to deduct the reason for it (although the definitions are made precise so you can solve this kind of problem by going back to your notes)
 
  • #8
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the answer is [1]
 
  • #9
HallsofIvy
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the answer is [1]
No, the answer is NOT (1)- you have been told that repeatedly!

One more time: the set [tex]\left[1, 3- \frac{1}{n}\right)[/tex], for any positive integer, n, is the set of all real numbers betwen 1 and 3- 1/n. For example, is n= 3, then [1, 3- 1/3) is the set of all real numbers between 1 and 3- 1/3= 8/3. It is NOT "an open set in Z" because it isn't even a set in Z!

You have also been told that (3) is not the correct answer. It is not an open set in this topological space.

This is getting ridiculous!
 
  • #10
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Z doesn't have a standard topology like R does. If we use the subspace topology, Z becomes the indescrete topology, and all sets are open (and also closed).

However, it's unlikely whatever text or teacher you're working with intended it this way. It's supposed to be a red herring answer. (To an uneducated student, it SEEMS reasonable, because there's a Z subscript in the set-builder notation).

In the standard (open ball-based) topology on R, the set [1,3-1/n) is neither open nor closed. The bracket [...) notation is a handy reminder of this fact. Intuitively, the reason it's not open is that open sets don't contain any boundaries, and the set contains 1, which is a boundary point.

In the subset [1,3), however, the point 1 is no longer a boundary point.

The correct answer is #2.

And this is basic definitions in point-set topology. Just because it boggled your mind doesn't mean it's "mind-boggling" :P
 
  • #11
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In the subset [1,3), however, the point 1 is no longer a boundary point.
But in Z the point 1 is no longer a boundary point also.
 
  • #12
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Note my comment about Z.

The question is definitely ambiguous. But if it's from a homework problem, the professor meant #1 to be an incorrect answer to throw off the students.
 
  • #13
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[tex]\left\{\left[1,3-\frac{1}{n}\right)\right\}_{n\in\mathbb Z} \text {is an "OPEN" cover of}\left[1,3\right)[/tex]



1.All [itex]\left[1,3-\frac{1}{n}\right)[/itex] is the open set in [itex]\mathbb Z[/itex] because [itex]\left(1,3-\frac{1}{n}\right)\cap\mathbb Z[/itex] where [itex]\left(1,3-\frac{1}{n}\right)[/itex] is open in R.

as i said before, of course [2] is valid but it is not the answer.

Recall: X, a collection of set, is said to be an open cover of [itex]S\subset K[/itex], if every set in X is open in [itex]K[/itex](not S!!!)
 
  • #14
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The question is definitely ambiguous.
that's why it is mind boggling
 
  • #15
Office_Shredder
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Recall: X, a collection of set, is said to be an open cover of [itex]S\subset K[/itex], if every set in X is open in [itex]K[/itex](not S!!!)
Are the intervals proposed open in R? The question seems like it might be ambiguous, but only one of the three options is factually accurate
 
  • #16
HallsofIvy
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Note my comment about Z.

The question is definitely ambiguous. But if it's from a homework problem, the professor meant #1 to be an incorrect answer to throw off the students.
that's why it is mind boggling
I see nothing ambiguous. The problem gives a collection of sets of real numbers indexed by integers. There is nothing in the problem that says anything about sets of integers.
 
  • #17
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I see nothing ambiguous. The problem gives a collection of sets of real numbers indexed by integers. There is nothing in the problem that says anything about sets of integers.
No, you probably wouldn't Halls :P

Speaking strictly, if we treat Z using the subset topology (where all sets are open) and take [1, 3-1/n) to be the set {1, 2} (the intersection of [1, 3-1/n) and the integers), you end up with a collection of open sets.

For a student who just learned about subset topologies, it's confusing. It's true on a technicality. It's not relevant to the open cover, and it's clearly (to us) a problem meant to distract the student from the right answer, but using totally reasonable choices of meanings for "Z" and "[1, 3-1/n)", [1,3-1/n) is in fact open in Z.
 
  • #18
disregardthat
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This reminds me of an anecdote from my teacher a few days ago: a student was having his oral topology exam, and was a nervous wreck in front of his examiners. He clearly had no firm grasp of the notion of compactness of spaces, so trying to be nice one of the professors asked if he could give an example of a compact space. The students answer was 'the real line'. The other professor, picking up on what the first one tried to do asked the follow-up question: "in what topology"?

The point here is that [tex][1, 3- \frac{1}{n})[/tex] can be written as an intersection between an open set in [tex]\mathbb{R}[/tex] and [1,3), hence it is open in [1,3) by the definition of the subspace topology. Pick for example the open subset [tex](0,3- \frac{1}{n})[/tex] of [tex]\mathbb{R}[/tex], and we have [tex][1, 3- \frac{1}{n}) =(0, 3- \frac{1}{n}) \cap [1,3)[/tex] open in [1,3). Therefore the union is an open cover, since [tex]3-\frac{1}{n} \to 3[/tex].

That the subspace topology is the relevant topology is clear from context. It should be clearly stated otherwise if not. I don't see what [tex]\mathbb{Z}[/tex] has to do with this, there are more than uncountably many subspace topologies of [tex]\mathbb{R}[/tex], so why [tex]\mathbb{Z}[/tex]?
 
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