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Union and Intersection of Open Sets (proof check) Watch

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    Doing the first two:



    (a) This one makes sense to me in the head though I'm not sure if I put it down on paper correctly and rigorously enough.

    Say A_1, A_2, ... \subset \mathbb{R}^n are open. As such, there must be some open ball in each A_i of radius \epsilon > 0 centered at any x \in A_i.

    Now consider A_i \cup A_j for some i \neq j then by the definition of a union, if B_{\epsilon}(x) \subset A_i then B_{\epsilon}(x) \subset \{ A_i \cup A_j \} - hence A_i \cup A_j is open.

    Similarly, we can extend this argument to infinity, hence \displaystyle \bigcup_{k=1}^{\infty} A_k is open.


    (b) Slightly more complicated but I gave it a go

    Say A_1, A_2, ... \subset \mathbb{R}^n are open. As such, there must be some open ball in each A_i of radius \epsilon > 0 centered at any x \in A_i.


    Now consider A_i \cap A_j for some i \neq j. If the intersection is \emptyset, then the set is open. (but is this valid? Since \emptyset is both open and closed)
    If the intersection is not \emptyset then there must be an open ball at points within the intersection, hence the intersection is open.

    How can I go about the the infinity part? As I can see it, intersections to infinity often leave you with an empty set which is both open and closed, so same argument as above with it? Not too certain on this.

    Came up with \displaystyle \bigcap_{n=1}^{\infty} \left( -\frac{1}{n},\frac{1}{n} \right) leaving you with a closed set for this though.
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    (Original post by RDKGames)
    Doing the first two:

    (a) This one makes sense to me in the head though I'm not sure if I put it down on paper correctly and rigorously enough.

    Say A_1, A_2, ... \subset \mathbb{R}^n are open. As such, there must be some open ball in each A_i of radius \epsilon > 0 centered at any x \in A_i.
    Sorry, but i struggle to make sense of that last statement.

    A given x must be in at least one of the A_i, but not necessarily in all of them, or even necessarily more than one of them.


    Now consider A_i \cup A_j for some i \neq j then by the definition of a union, if B_{\epsilon}(x) \subset A_i then B_{\epsilon}(x) \subset \{ A_i \cup A_j \} - hence A_i \cup A_j is open.
    :holmes: Well you've chosen x in A_i, and whilst it's true B_\epsilon (x)\subset \{A_i\cup A_j\}, it doesn't necessarily follow that this union is open without considering x in A_j

    Similarly, we can extend this argument to infinity, hence \displaystyle \bigcup_{k=1}^{\infty} A_k is open.
    Well that doesn't necessarily follow. Just because a finite union is open, doesn't imply an infinite union is open.

    It's akin to saying \displaystyle\sum _{i=1}^ni is finite for all n, therefore \displaystyle\sum _{i=1}^\infty i is finite.

    HOWEVER, the biggest stumbling block to this inductive style proof, is that an infinite union does not necessarily imply a countably infinite union, so induction won't work.

    With this proof you need to grasp the whole nettle to start.

    Suppose we have a union of subsets of \mathbb{R}^n, indexed by some set I. This covers the finite, and infinite (countable and otherewise), possibilities. And let x be an element of this union. So.

    \displaystyle x\in\bigcup_{i\in I}A_i

    It follow that x must be in A_i for some i\in I

    Hence there is a ball....

    Can you take it from there - should only take a couple of lines.
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    (Original post by ghostwalker)
    Sorry, but i struggle to make sense of that last statement.

    A given x must be in at least one of the A_i, but not necessarily in all of them, or even necessarily more than one of them.
    I was just trying to say that there is an open ball in each A_i, so maybe saying x_i \in A_i is more accurate?


    :holmes: Well you've chosen x in A_i, and whilst it's true B_\epsilon (x)\subset \{A_i\cup A_j\}, it doesn't necessarily follow that this union is open without considering x in A_j
    So let me rephrase by orginal statement to align with the above, by saying that 'Let's pick some x \in A_i ... then B_{\epsilon}(x) \in \{ A_i \cup A_j \}" but this has to clearly be open, I'm not sure how to make it more clear? x doesn't have to be in A_j, but even if it is, since A_j is open then the ball around x is open, hence the union is still open as it contains the ball. I'm not sure what you mean by considering x in A_j then, if it's of any significance more than this.

    Well that doesn't necessarily follow. Just because a finite union is open, doesn't imply an infinite union is open.

    It's akin to saying \displaystyle\sum _{i=1}^ni is finite for all n, therefore \displaystyle\sum _{i=1}^\infty i is finite.
    Not sure how to make it follow from the above then.

    HOWEVER, the biggest stumbling block to this inductive style proof, is that an infinite union does not necessarily imply a countably infinite union, so induction won't work.

    With this proof you need to grasp the whole nettle to start.

    Suppose we have a union of subsets of \mathbb{R}^n, indexed by some set I. This covers the finite, and infinite (countable and otherewise), possibilities. And let x be an element of this union. So.

    \displaystyle x\in\bigcup_{i\in I}A_i

    It follow that x must be in A_i for some i\in I

    Hence there is a ball....

    Can you take it from there - should only take a couple of lines.
    Hence there is a ball in A_i around x, which means that A_i is open and also the union is open?
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    (Original post by RDKGames)
    I was just trying to say that there is an open ball in each A_i, so maybe saying x_i \in A_i is more accurate?
    I'm sorry, but I'm not clear what you're saying / trying to say, so don't know how to respond. And to be honest, it's beside the point, as this line of proof won't work, since you can't jump from the finite case to an infinite one.

    Hence there is a ball in A_i around x, which means that A_i is open and also the union is open?
    Well you know A_i is open, as that's a given.

    To show that the union is open, you want to show that given any x in the union, then there is an open ball centred on x that is a subset of the union.

    Last post for today, so I'll cut to the chase - here's the outline:

    Spoiler:
    Show


    Let x be in the union.

    Then x is in A_i from some i.

    Since A_i is open, then there is an open ball B_\epsilon (x)\subset A_i

    NOTE: Epsilon will depend on the choice of x.

    Since A_i\subset \bigcup A_i

    it follows that B_\epsilon (x)\subset \bigcup A_i

    and hence the union is open.

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    (Original post by ghostwalker)
    I'm sorry, but I'm not clear what you're saying / trying to say, so don't know how to respond. And to be honest, it's beside the point, as this line of proof won't work, since you can't jump from the finite case to an infinite one.



    Well you know A_i is open, as that's a given.

    To show that the union is open, you want to show that given any x in the union, then there is an open ball centred on x that is a subset of the union.

    Last post for today, so I'll cut to the chase - here's the outline:

    Spoiler:
    Show




    Let x be in the union.

    Then x is in A_i from some i.

    Since A_i is open, then there is an open ball B_\epsilon (x)\subset A_i

    NOTE: Epsilon will depend on the choice of x.

    Since A_i\subset \bigcup A_i

    it follows that B_\epsilon (x)\subset \bigcup A_i

    and hence the union is open.



    Thanks, gonna have a go at doing the intersection one again in the morning as I'd imagine it follows the same pattern as this one.

    Seems I've lost the little amount of analysis skills that I had over the summer :ashamed2:
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    (Original post by RDKGames)
    (b) Slightly more complicated but I gave it a go
    Ghostwalker has covered the first bit thoroughly, so:

    Say A_1, A_2, ... \subset \mathbb{R}^n are open. As such, there must be some open ball in each A_i of radius \epsilon > 0 centered at any x \in A_i.
    You're doing this for finitely many subsets, so you should be considering A_1, A_2, \cdots, A_m \subset \mathbb{R}^n.

    Now consider A_i \cap A_j for some i \neq j. If the intersection is \emptyset, then the set is open. (but is this valid? Since \emptyset is both open and closed)
    If the intersection is not \emptyset then there must be an open ball at points within the intersection, hence the intersection is open.
    Not entirely sure what you're doing here or why you're considering pairwise intersections. You want to show that A = \displaystlye \bigcap_{j=1}^{m} A_j is open. Take arbitrary x\in A then x\in A_j for all 1 \leqslant j \leqslant m. So you can find an open ball B(x, \epsilon_j)\subset A_j for every 1 \leqslant j \leqslant m.

    Now can you pick an \epsilon such that B(x, \epsilon) \subset B(x, \epsilon_j) \subset A for every 1 \leqslant j \leqslant m?

    [hint, you only have finitely many \epsilon_j's, so there's something you can pick from them]

    How can I go about the the infinity part? As I can see it, intersections to infinity often leave you with an empty set which is both open and closed, so same argument as above with it? Not too certain on this.

    Came up with \displaystyle \bigcap_{n=1}^{\infty} \left( -\frac{1}{n},\frac{1}{n} \right) leaving you with a closed set for this though.
    Not really sure what you're asking. You want to show that infinite intersections of open sets are not necessarily open. This means you need to provide a single counterexample, not a "same argument as above".

    Your example is fine: each (-1/n, 1/n) is open but their infinite intersection is \{0\} which is not open [it is not enough to say that it is closed, remember that a set can be both closed and open]
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    (Original post by RDKGames)
    ..
    Not to pile on, but it seems to a major problem you're having is not knowing where/how to start.

    And in fact, this isn't something that's difficult, it's essentially "boilerplate" (like how in first year analysis everything starts with "take epsilon > 0..." )

    If you have a set E that you want to show is open, then 99% of the time, your approach is going to be

    "Pick x \in E, then {somehow} show we can find \epsilon> 0 s.t. B_\epsilon(x) \subseteq E".
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    OK, some good tips up there, thanks guys, so I had another go at the intersections one:

    Let \displaystyle A=\bigcap_{i=1}^n A_i, thus A \subset \mathbb{R}^n. Pick x \in A then we have x \in A_1, x=A_2, ..., x \in A_n. Since each A_j is open, for 1 \leq j \leq n, then there exists B_{\epsilon_j}(x) \subset A_j. Picking \epsilon = \min{\epsilon_j}, we can say that there is an open ball B_{\epsilon}(x) which is contained in every A_j; as such B_{\epsilon}(x) \subset B_{\epsilon_j}(x) \subset A_j \subset A

    Thus B_{\epsilon}(x) \subset A hence A is open.

    \displaystyle \bigcap_{i=1}^{\infty} A_i is not always open as per my example of A_i=(-\frac{1}{i},\frac{1}{i})




    Then had a go at (c):

    To show that A=\displaystyle \bigcap_{m=1}^p A_m is closed, where A_k is closed for 1 \leq k \leq p, we need to show that A^c=\mathbb{R}^n / A is open.

    Let's note that by De Morgan's laws we have \displaystyle A^c= \left( \bigcap_{m=1}^p A_m \right)^c = \bigcup_{m=1}^p A_m^c

    We have A^c_k being open since A_k is closed. Then we have a union of finite open sets which is open as per part (a). Thus we have A^c being open, hence A is closed.

    It also follows from (a) that \displaystyle \bigcap_{m=1}^{\infty} A_m is also closed.

    I didn't go about the infinity part correctly in (a) so here's my attempt for it returning to (a)

    Say \displaystyle A=\bigcup_{i=1}^{\infty}A_i where A_k is open for k \in \mathbb{N}. Picking x \in A means x \in A_k for some k. As such, since A_k is open we have B_{\epsilon}(x) \subset A_k \subset A thus A is open.
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    (Original post by RDKGames)
    ..
    I'm not quite sure what you're trying to show in (c); the question explictly says you need to prove even an intersection of an infinite number of closed sets is closed, and yet you've made a proof for a finite intersection. Your "it also follows trom (a) that the infinite intersection is closed" has no justification and is unconvincing.

    That said, this is largely a niggle - your argument works just as well for an infinite intersection, just use the appropriate notation instead of \bigcap_{i=1}^p.

    A second point that is kind of a niggle, but also important: in analysis / topology, you should not assume that an infinite union or intersection is countably infinite. You'll notice Ghostwalker used an indexing set I, rather than assuming the union could be indexed 1, 2, ..., (to infinity). It's not at all uncommon to do something like have an particular set S, and make an open set O containing it by writing \displaystyle O = \bigcup_{x\in S} B_{\epsilon_x}(x), (where \epsilon_x will depend on x and you've given some argument about how to decide what \epsilon_x is), and you want to know this is open even if S is uncountable.
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    (Original post by DFranklin)
    I'm not quite sure what you're trying to show in (c); the question explictly says you need to prove even an intersection of an infinite number of closed sets is closed, and yet you've made a proof for a finite intersection. Your "it also follows trom (a) that the infinite intersection is closed" has no justification and is unconvincing.

    That said, this is largely a niggle - your argument works just as well for an infinite intersection, just use the appropriate notation instead of \bigcap_{i=1}^p.
    Yes it asks for infinite but also finite, from the way I've read it. So I've proved the finite case and said that the infinite one follows from the very similar argument so I wouldn't have to write the same things again.

    So for the infinite+finite intersection it would go like this?

    Let \displaystyle A = \bigcap_{i \in I} A_i be the intersection of closed sets A_i in \mathbb{R}^n and indexed by some set I. To show this is closed, we show that A^c=\mathbb{R}^n / A is open.

    From De Morgan's laws we have \displaystyle A^c = \left( \bigcap_{i \in I}A_i \right)^c=\bigcup_{i \in I} A_i^c

    Now since A_k is closed, for some k \in I, then A_k^c is open. (then this is just followed from part a's proof for union of open sets being open)

    Pick x \in A_k^c then there is a ball B(x,\epsilon) in A_k of an appropriate radius \epsilon>0.

    Since \displaystyle B(x,\epsilon)\subset A_k \subset \bigcup_{i \in I} A_i^c then we have an open ball in the union, thus A^c is open, and A is closed.
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    (Original post by RDKGames)
    So for the infinite+finite intersection it would go like this?

    Let \displaystyle A = \bigcap_{i \in I} A_i be the intersection of closed sets A_i in \mathbb{R}^n and indexed by some set I. To show this is closed, we show that A^c=\mathbb{R}^n / A is open.
    The standard notation is \mathbb{R}^n\setminus A. I've never seen \mathbb{R}^n / A used. "\setminus" in LaTeX.


    From De Morgan's laws we have \displaystyle A^c = \left( \bigcap_{i \in I}A_i \right)^c=\bigcup_{i \in I} A_i^c

    Now since A_k is closed, for some k \in I, then A_k^c is open. (then this is just followed from part a's proof for union of open sets being open)

    Pick x \in A_k^c then there is a ball B(x,\epsilon) in A_k^c of an appropriate radius \epsilon>0.

    Since \displaystyle B(x,\epsilon)\subset A_k^c \subset \bigcup_{i \in I} A_i^c then we have an open ball in the union, thus A^c is open, and A is closed.
    Be careful with the use of "for some". Although I know what you intend, this wording means, "there exists" or "there is at least one". You want "for any", or "for all".

    I've modified your last couple of lines to include the complement sign on two occasions. I assume they were typos.

    I would not have repeated details from part (a), and instead just say:

    Since A_k^c is open for all k \in I, then by (a), \displaystyle\bigcup_{k\in I} A_k^c is open, etc.

    There is, of course, no harm in adding the additional detail. BUT if you do, you want to start from x\in\bigcup A_k^c (which is stating that you're dealing with a general element in the union), And go on, "so x\in A_k^c for some k in I". And continue as you did.

    Again, small points, but not insignificant.
 
 
 
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