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 are open. As such, there must be some open ball in each of radius centered at any .
Now consider for some then by the definition of a union, if then  hence is open.
Similarly, we can extend this argument to infinity, hence is open.
(b) Slightly more complicated but I gave it a go
Say are open. As such, there must be some open ball in each of radius centered at any .
Now consider for some . If the intersection is , then the set is open. (but is this valid? Since is both open and closed)
If the intersection is not 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 leaving you with a closed set for this though.

RDKGames
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 11102017 21:51
Last edited by RDKGames; 11102017 at 21:55. 
ghostwalker
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 11102017 22:37
(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 are open. As such, there must be some open ball in each of radius centered at any .
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.
Well you've chosen x in A_i, and whilst it's true , it doesn't necessarily follow that this union is open without considering x in A_j
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 is finite for all n, therefore 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 , 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.
It follow that x must be in for some
Hence there is a ball....
Can you take it from there  should only take a couple of lines.Last edited by ghostwalker; 11102017 at 22:41. 
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 11102017 23:27
(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.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 , 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.
It follow that x must be in for some
Hence there is a ball....
Can you take it from there  should only take a couple of lines.Last edited by RDKGames; 11102017 at 23:30. 
ghostwalker
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 12102017 00:00
(Original post by RDKGames)
I was just trying to say that there is an open ball in each , so maybe saying is more accurate?
Well you know 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:

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 12102017 00:04
(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 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:
Seems I've lost the little amount of analysis skills that I had over the summer 
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 12102017 01:06
(Original post by RDKGames)
(b) Slightly more complicated but I gave it a go
You're doing this for finitely many subsets, so you should be considering .
Now can you pick an such that for every ?
[hint, you only have finitely many 's, so there's something you can pick from them]
Your example is fine: each is open but their infinite intersection is which is not open [it is not enough to say that it is closed, remember that a set can be both closed and open]Last edited by Zacken; 12102017 at 01:09. 
DFranklin
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 12102017 09:31
(Original post by RDKGames)
..
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 , then {somehow} show we can find s.t. ". 
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 12102017 14:17
OK, some good tips up there, thanks guys, so I had another go at the intersections one:
Let , thus . Pick then we have , , ..., . Since each is open, for , then there exists . Picking , we can say that there is an open ball which is contained in every ; as such
Thus hence is open.
is not always open as per my example of
Then had a go at (c):
To show that is closed, where is closed for , we need to show that is open.
Let's note that by De Morgan's laws we have
We have being open since is closed. Then we have a union of finite open sets which is open as per part (a). Thus we have being open, hence is closed.
It also follows from (a) that 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 where is open for . Picking means for some . As such, since is open we have thus A is open.Last edited by RDKGames; 12102017 at 14:19. 
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 12102017 14:52
(Original post by RDKGames)
..
That said, this is largely a niggle  your argument works just as well for an infinite intersection, just use the appropriate notation instead of .
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 , (where will depend on x and you've given some argument about how to decide what is), and you want to know this is open even if S is uncountable. 
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 12102017 19:52
(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 .
So for the infinite+finite intersection it would go like this?
Let be the intersection of closed sets in and indexed by some set . To show this is closed, we show that is open.
From De Morgan's laws we have
Now since is closed, for some , then is open. (then this is just followed from part a's proof for union of open sets being open)
Pick then there is a ball in of an appropriate radius .
Since then we have an open ball in the union, thus is open, and is closed.Last edited by RDKGames; 12102017 at 19:54. 
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 12102017 21:56
(Original post by RDKGames)
So for the infinite+finite intersection it would go like this?
Let be the intersection of closed sets in and indexed by some set . To show this is closed, we show that is open.
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 is open for all , then by (a), is open, etc.
There is, of course, no harm in adding the additional detail. BUT if you do, you want to start from (which is stating that you're dealing with a general element in the union), And go on, "so for some k in I". And continue as you did.
Again, small points, but not insignificant.
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