<Ozera> if two vertices are connected by an edge, do I call the vertices adjacent?
<hunteriam> http://imgur.com/2IkokOd
<hunteriam> can anyone help me with this question
<hunteriam> ive made some progress into it but im now certainly stuck
<bollu> Galois: yes, Rudin's trying to define when a set will be open in a subspace
<bollu> Galois: but it doesn't make sense. Like, G could be open in X, but I don't see why it needs to be open in Y
<bollu> for example, if X = [0, 10], Y = [0, 1], G = (0.5, 1.5)
<Galois> ok, so you're using metric spaces
<bollu> G is open in X
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<bollu> G intersection Y is _not_ open in Y
<Galois> why not?
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<bollu> G intersection Y is (0.5, 1]
<Galois> so?
<bollu> that's neither open nor closed in [0, 1] right?
<Galois> huh?
<Galois> use the definition of open in Y
<Galois> which is why I asked you
<Galois> what's the definition?
<lethargicasd> when a matrix has entries with the exact same value per column, what does this imply? eg {{2 1 1}{2 1 1}{2 1 1}}
<bollu> every point is an interior point
<Galois> name a point of (0.5, 1] that's not an interior point
<hunteriam> lethargicasd: what sort of things are you looking for it to imply?
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<bollu> how is 1 an interior point?
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<Galois> what's the definition of interior point?
<bollu> it
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<bollu> an interior point "p" is a point such that there exists an open ball V(p, r) that completely belongs to the set
<lethargicasd> hunteriam im looking to see what it associates with eigenvectors/eigenvalues/rank
<Galois> ok, what's the definition of open ball?
<hunteriam> lethargicasd: it means the dot product of that matrix and any vector is going to be a vector of the form {a,a,a}
<bollu> an open ball is just the set of all points V(p0, r) =  {x, |x - p0| < r}
<tmg_> lethargicas, check out rank--nullity theorem
<Galois> almost, so close
<Galois> An open ball in Y is the set V(p0, r) =  {x in Y, |x - p0| < r}
<bollu> ohhh
<bollu> so your "open ball" is a "ball" squeezed in the space
<Galois> if you're not super ultra strict with your definitions you will lose like this
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<Galois> and indeed, "{x, |x - p0| < r}" is meaningless notation
<bollu> so, here, for 1, the open ball could be something like V(p0=1, r=0.1) = (0.9, 1) ?
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<bollu> right, I missed the |
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<bollu> :)
<Galois> you missed the in Y
<Galois> that's required under set theory
<bollu> oh, to say where everything comes form
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<Galois> I forgave the typo
<bollu> from*?
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<Galois> in any case, V(p0=1, r=0.1) is (0.9,1]
<Galois> not (0.9,1)
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<bollu> ah, okay. The point 1 also gets included 'case it's at distance 0?
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<bollu> 'cause*
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<bollu> okay, so, in that case, G intersection Y _is_ open in Y?
<bollu> Galois: okay, I got an idea to prove that. Can you verify it for me?
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<bollu> consider Y, a subset of X, and G that is open in Y
<Galois> you can build a set without "in Y" on the left, but you would need an "in Y" on the right. For example: {2*x : x in X} is valid.
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<Galois> if you really care about the ugly details, look at the axiom (schema) of specification and the axiom (schema) of replacement. Otherwise just understand the basic requirement that with very few exceptions (e.g. the axiom that says "there exists a set"), you can only build a set from an existing  set
<tjt263> Quantumplation:
<tjt263> for instance i understand “exponential” to mean something essentially like a snowball effect accumulating more and more quicker and quicker because;
<tjt263> as the snowballs gets bigger, the increased surface area allows it to get bigger (and heavier), quicker.
<tjt263> the bigger (and heavier)that it gets; the faster it gets bigger
<tjt263> and so on, like a positive feedback loop right?
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<bollu> write G as (union (i = 0 to inf) Y_i) union (union (j = 0 to inf) X_i) where X_i are open sets of (X intersection Y complement), Y_i are open sets of Y. so, now, G intersection Y will just be all the Y_i, which were defined to be open in Y.
<bollu> Galois: oh, so axiomatic set theory forces you to build sets from other sets?
<Galois> yes
<bollu> I think my definition of G works out to be tautological in this case
<bollu> I need to prove that you _can_ write G that way
<Galois> I'm not convinced by your proof, because: how do you know the unions are countable?
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<bollu> Galois: they don't need to be, right? inifite union still works?
<bollu> infinite*
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<bollu> oh, does it have to be countably infinite?
<Galois> ok, but that's not what you wrote
<Galois> if you didn't write it, it doesn't count
<bollu> i = 0 to inf ?
<Galois> no good, 0 to inf is a countable set
<bollu> ohh
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<bollu> hm. How do you denote uncountable union then?
<bollu> i = 0 to 2^N? :)
<x86_64> if i do this: 5^23 , what is the verb? Am i ¨raising¨ 5 to 23? how do i say my action in english?
<Galois> you can't just arbitrarily write an uncountable union. You need a set over which to index.
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<Galois> "0 to inf" implies you're indexing over the set N of natural numbers
<bollu> ahh. Hm, I choose to index over all real numbers from [0, 1] ?
<bollu> Galois: I don't know, I've never had to formally use uncountable union before
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<Galois> that's horribly complicated. index over the elements of G
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<bollu> Galois: could you show me how to do that? or.. is there an alternate way to prove what I was trying to prove?
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<bollu> Galois: but, I'm constructing G. I can't index over it
<Galois> you're proving B => A. You're given G.
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<phc> lol
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<bollu> Um, okay. Can we take a step back?
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<bollu> new question :)
<bollu> I'm given Y is a subset of X. I know G is open in X
<bollu> I want to prove that G intersection Y is open in Y
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<waterCreature> hi, i am doing simplex method and I am at the step to find the pivot. I  am following this guide to simplex method.pdf from wsu and it says I choose pivot column the most negative indicator
<Galois> OK, so now you're asking about A => B
<bollu> Galois: yes
<Galois> that's not what you asked in your original question, but ok
<bollu> Galois: I thought I understood A => B but our discussion proved I didn't >_<
<waterCreature> and to select the pivot row, I need to divide the RHS with the non-zero entry of each row.
<waterCreature> Does that include negative values?
<bollu> okay, so, now
<Galois> I think A => B is just definition chasing
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<bollu> I can say, all p in G is an interior point of G. So, there will be an open ball V(p, r) for some r > 0 such that V(p, r) belongs to G. Now, for every p, consider V(p, r) intersection Y
<Galois> you need to prove that "V(p,r) interpreted in Y" is equal to "V(p,r) intersect Y"
<bollu> yeah, I need to show that V(p, r) intersection Y remains open in Y
<Galois> and I think that's only true if p is in Y
<bollu> ah, right!
<Galois> if p is in G but not in Y, you can't really say anything
<bollu> right. So pick all the p that are common in G and in Y.
<bollu> then take the neighbourhoods and union them up
<Galois> there are two different V(p, r)'s here. In X: V(p, r) = {x in X : d(x,p) < r}
<Galois> In Y: V(p, r) = {y in Y : d(y,p) < r}
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<bollu> won't (V(p, r) in X intersection Y) be (V(p, r) in Y)?
<Galois> yes, but you need to prove that
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<bollu> okay, that seems doable. Consider Y, the subset of X. construct the open ball V_X(p, r) = {x in X | d(x, p) < r }. Now, consider V_Y(p, r) = {y in Y | d(y, p) < r }. all V_y will belong to V_x
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<bollu> and V_y will contain only the elements that are common to X and Y (since Y is a subset of X and I'm picking form Y)
<bollu> picking from*
<Galois> much too complicated. This question is about set theory, not topology.
<bollu> hm, right
<Galois> "Suppose z is an element of ((V(p, r) in X) intersection Y). Then d(z,p) < r [definition of (V(p,r) in X)], and z is in Y [definition of intersection]. Hence z is in (V(p, r) in Y)."
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<Galois> "Therefore ((V(p, r) in X) intersection Y) is a subset of (V(p,r) in Y)"
<Galois> that's one direction. You do the other direction.
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<bollu> "Suppose z is an element of (V_Y(p, r)). Then, pick V_X(p, r). z will also be an element of V_X(p, r) (since z is an element of Y which is a subset of X). Now, since z is also an element of Y, z will belong to (V_X(p, r) intersection Y)
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<Galois> you never used the defining property of V(p,r), which is that d(z,p) < r
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<bollu> Then, pick V_X(p, r). z will also be an element of V_X(p, r) (since z is an element of Y which is a subset of X | I'm using it implicitly here when talking about the open balls, aren't I?
<Galois> you should use it explicitly. Since the entire fact you're proving is something about open balls, it is circular to use open balls implicitly.
<bollu> ah
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<bollu> Okay, second stab at it
<Galois> when you get good, you can skip these housekeeping steps. But right now you should not skip them, or else you'll get into trouble, just like you did when you thought that 1 was not an interior point of [0,1]
<bollu> Galois: so, just like epsilon-deltamanship :)
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<bollu> Given: (V(p, r) is open in Y) To prove: (V(p, r) is open in X intersection Y)
<Galois> uh, noooooooooooooooooooooooooo
<bollu> Galois: no?
<bollu> >_<
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<Galois> you simply have to prove that V(p,r) (as defined in Y) is equal to ((V(p,r) (as defined in X)) intersect Y)
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<Galois> you're not tryint to prove that V(p,r) is open in X or open in anywhere
<bollu> ohh
<Galois> "open" is a property of V(p,r)
<Galois> this is not what you're trying to prove
<Galois> you're trying to prove that "This set" is equal to "This other set"
<bollu> Galois: okay, so we're trying to prove "set theoretic equivalence" given the extra structure of the topology?
<bollu> ahh, gotcha
<Galois> not that "This set" has "This other property"
<Galois> there's no topology here at all. You're trying to prove that "This set" is equal to "This other set"
<Galois> do you know the definition of set equality?
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<bollu> Galois: yeah, A is a subset of B and B is a subset of A
<bollu> then A = B
<Galois> ok, and notice that I proved A is a subset of B
<abcdef_guy> Does W^{1,1}([0,1]) embed into C([0,1])?
<bollu> right, you picked an element from A and showed that it'll always be in B
<bollu> where A = V_X(p, r) intersection Y, and B = V_Y(p, r)
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<bollu> okay, lemme try again: Given: (z is in V_Y(p, r) To prove: (z is in (V_X(p, r) intersection Y))
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<bollu> pick z in V_Y(p, r). This implies that |z - p| < r for p belongs to Y. However, both z, p also belong to X (since Y is a subset of X). So, the statement |z - p| < r also holds in X. Therefore, z is an element of V_X(p, r). However, z is also from Y, so z is an element of V_X(p, r) intersection Y
<bollu> Galois: how was that?
<waterCreature> and to select the pivot row, I need to divide the RHS with the non-zero entry of each row.
<waterCreature> hi, i am doing simplex method and I am at the step to find the pivot. I  am following this guide to simplex method.pdf from wsu and it says I choose pivot column the most negative indicator
<waterCreature> and to select the pivot row, I need to divide the RHS with the non-zero entry of each row.
<waterCreature> Does that include negative values?
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<waterCreature> like in one of the rows, the non-zero entry value is -1
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<Galois> better. I think most instructors would accept that as valid.
<waterCreature> If i divided the last column with -1, I would get -4
<waterCreature> and that would be the smallest
<bollu> cool :). Okay, so now we're armed with the fact that V_Y(p, r) == V_X(p, r) intersection Y given Y is a subset of X
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<waterCreature> so.. that row would become the pivot row.?
<Ozera> For a complete graph K_n, K_{n-1} is a subgraph of K_n right?
<bollu> Ozera: yep, drop the extra point and all edges wth that point in K_n
<abcdef_guy> Anybody?
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<Ozera> bollu, right -- thanks
<abcdef_guy> I think the answer is yes because W^(1,1)([0,1]) functions are absolutely continuous
<bollu> Galois: now, we were trying to prove that, if, G is open in X, then G intersection Y will be open in Y, right?
<Galois> abcdef sorry i don't know functional analysis
<bollu> given Y is a subset of X
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<bollu> Galois: so, now, I was saying, for every g in G, pick the neighbourhood V_X(g, r) that is in X. You'll always have such a neighbourhood since G is open in X
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<bollu> Galois: now, construct V_X(g, r) intersection Y. this is the same as picking up V_Y(g, r), right?
<bollu> as we proved?
<Galois> so far so good
<bollu> "the same"  = "the same set"
<yyu43> Is a finitely generated submodue of a free module always free?
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<bollu> since V_X(g, r) was open in X, consider V_Y(g, r). This consists of all points z such that |z - g| < r. So, V_Y is open in Y as well
<Galois> http://mathoverflow.net/questions/16953/are-submodules-of-free-modules-free ???
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<bollu> therefore, G can be written as the union (for all g belongs to G) V_Y(g, r)
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<bollu> Galois: this completes the proof?
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<Galois> you still haven't proved that V_Y(g,r) is a subset of G
<bollu> grr
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<Goldbach> is it true that galois read math books like novels and mastered at first reading?
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<Goldbach> for him, math was surely a spectator sport as evidenced by his contemporaries
<Goldbach> for all the greats, math is indeed a spectator sport
<Ozera> Galois, do tell us please
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<Galois> you actually have to prove that V_Y(g,r) is a subset of (G intersect Y), but I take it as "obvious" that V_Y(g,r) is a subset of Y, so you only have to prove the "subset of G" part
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<bollu> Galois: hm, right
<Galois> I don't think they had math books in 1823
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<Goldbach> well.. yes, they had articles
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<bollu> okay, since G == Union (for all g in G) V_X(g, r) || G_Y = Union (for all g in G) V_Y(g, r). G_Y wil be a subset of G (since every V_Y(g, r) is a subset of V_X(g, r)).
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<bollu> Therefore, G_Y is a subset of G, which is open in Y (since it was constructed form open sets in Y)
<bollu> where G_Y = G intersect Y
<bollu> (by construction)
<Galois> I'm not even going to read that, because even if it's right, it's far too complicated, and thus is "morally" wrong
<bollu> aww
<bollu> okay
<bollu> consider this
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<bollu> G is open in X
<bollu> check?
<bollu> now consider G intersect Y
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<bollu> I can break G into V(g_1, r_1) union V(g_2, r_2) … union V(g_n, r_n)
<Galois> bzzt. too complicated.
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<bollu> Galois: :( I was going to distribute intersection
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<Galois> I'm going to give up now, since I have no more time.
<Galois> it's like one sentence:
<bollu> do tell :)
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<Galois> V_Y(g,r) = V_X(g,r) intersect Y; in particular, V_Y(g,r) is a subset of V_X(g,r), and V_X(g,r) is a subset of G by construction of V_X(g,r), so V_Y(g,r) is a subset of G
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<Galois> and now I see a mistake in your proof above: "for every g in G, pick the neighbourhood V_X(g, r) that is in X"
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<Galois> it should have said: "for every g in G, pick the neighbourhood V_X(g, r) that is in G"
<bollu> ah, right :) It should be interior.
<bollu> whoaa
<bollu> so you just nested V_Y(g, r) into V_X(g, r) into G
<Galois> and this is why you always should do the proofs so carefully: because it makes it easy to catch mistakes
<bollu> Galois: Topology is the first-ish subject I'm maintaining a notebook for, because the steps are so.. precise
<Galois> if you were doing graph theory or algebra or whatever correctly, it should also be as precise
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<bollu> Galois: yes, but they're more "obvious" to see, I guess. Being geometric / visualisable
<Galois> it's just that you can get away with it in those subjects. Well, you can't here.
<bollu> yeah… I'll try and do this from scratch again, paying attention :)
<bollu> thanks a ton for all the help!
<Galois> when I teach algebra I insist on absolute rigor and precision, even though you can get away without it, because it's easier to learn good habits in a simpler setting
<Galois> as opposed to now where you're thrust into metric spaces and you have to learn new (good) habits while simultaneously learning difficult content
<bollu> yeah :) But I'm enjoying the experience!
<Ozera> bollu, is there any particular notation to denote H being a subgraph to G ?
<Ozera> of G*
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<bollu> Ozera: none that I'm aware of, no. My algebraic graph theory book just uses subset notation and gets away with it :P
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<Ozera> oooooh *algebraic* graph theory?
<Ozera> that sounds fun
<bollu> Ozera: Yep! take group theory, and use it to study graph theoretic properties like cut sets, adjacencies and whatnot.
<Ozera> seet
<Ozera> sweet*
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<bollu> does #math maintain history? I closed my IRC client by mistake
<Naenil> You could buy a bouncer but i'm sure someone logs
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<Mufleta> someone asked for that yesterday and people said there isnt one.   some ppl here tho keep their own log
<Ozera> I keep my own log
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<bollu> Ozera: could you dump the past half hour for me on pastebin or something?
<Ozera> uh