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Topology and Continuity question

  1. Oct 5, 2005 #1

    AKG

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    Let [itex]\mathbb{R}_{l}[/itex] denote the real numbers with the lower limit topology, that is the topology generated by the basis:

    [tex]\{[a, b)\ |\ a < b,\ a, b \in \mathbb{R}\}[/tex]

    Which functions [itex]f : \mathbb{R} \to \mathbb{R}[/itex] are continuous when regarded as functions from [itex]\mathbb{R}_l[/itex] to [itex]\mathbb{R}_l[/itex]? I have the following theorem:

    Let X, Y be topological spaces; let [itex]f : X \to Y[/itex]. Then the following are equivalent:

    a) f-1(U) is open in X for every open subset U of Y (definition of continuity).
    b) For every subset A of X, [itex]f(\overline {A}) \subset \overline{f(A)}[/itex].
    c) For every closed set B of Y, the set f-1(B) is closed in X.
    d) For each [itex]x \in X[/itex] and each neighbourhood V of f(x), there is a neighbourhood U of x such that [itex]f(U) \subset V[/itex].


    I'm not sure how to go about doing this.
     
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  3. Oct 5, 2005 #2

    StatusX

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    Think about the epsilon delta definition of continuity, and see how much of it can be salvaged in this topolgy.
     
  4. Oct 5, 2005 #3

    AKG

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    Let [itex]\mathbb{R}[/itex] denote the set of real numbers with the usual topology, that being the one generated by the set of open intervals:

    [tex]\{(a, b)\ |\ a < b;\ a,b \in \mathbb{R}\}[/tex]

    Again, [itex]\mathbb{R}_l[/itex] denotes the reals with the lower-limit topology. A function from [itex]\mathbb{R}_l[/itex] to [itex]\mathbb{R}_l[/itex] is continuous if and only if, when treated as a function from [itex]\mathbb{R}[/itex] to [itex]\mathbb{R}[/itex], it is:

    a) continuous from the right, i.e. [itex]\lim _{x \to a^+} f(x) = f(a)[/itex] for all real a, and
    b) non-decreasing on intervals where it is continuous.

    If condition b) doesn't hold, then if we take for example f(x) = -x, then f-1([0, 1)) = [-1, 0) which isn't open. If the function is continuous as a function on [itex]\mathbb{R}_l[/itex], then for every x and every e > 0, [f(x), f(x) + e) is an open set and hence its preimage will be some open set containing x, and thus must contain [x, x + d) for some d so f is continuous from the right. It remains to show that if a) and b) hold, then f is continuous on [itex]\mathbb{R}_l[/itex]. If V is open in the co-domain, we'd like to show that its preimage is a union of [x, x+d) for each x in the preimage of V. Well we know that for every f(x) in V, there is an e such that [f(x), f(x) + e) is contained in V since V is open, and for every e, there is a d such that f([x, x + d)) is contained in (f(x) - e, f(x) + e) since f is right-continuous. In fact, if we choose d such that f is continuous on [x, x+d) then we know that not only is f([x, x+d)) contained in (f(x) - e, f(x) + e), but that it can contain no element less than f(x) by condition b) so it is contained in [f(x), f(x) + e) which is in turn contained in V.

    Does this proof look complete? I know the first sentence is a little wishy-washy since it only gives one example f(x) = -x and doesn't prove it in general, but I think that should be easy.
     
  5. Oct 6, 2005 #4

    StatusX

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    That sounds good. You don't have to treat the function as one from R to R; the limit definition depends only on distance, not topology, and is equaly valid in any topology of R. And make sure you prove that not only do the two conditions imply the function is continuous, but also the continuity of the function should imply each of the two conditions. You've proved two of these three directions, and mentioned you'll prove the last one later, so you should be fine.
     
  6. Oct 6, 2005 #5

    AKG

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    My proof above is actually missing stuff, and I don't even think the proposition is correct, but I did something else which I believe is correct. Thanks anyways.
     
  7. Oct 6, 2005 #6

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    I think your original assertion is correct. Here's how I would prove the last part.

    Assume f is continuous in Rl. Then for every x in the domain, there is some half open interval in the range [f(x),f(x)+e), with e a positive number, and the preimage of this interval is open. Then [x,x+d) belongs to this preimage for some positive d. Assume the function is decreasing at x. Then [x,x+d) does not map into [f(x),f(x)+e) for any positive number d, contradicting the previous statement. Thus the function is non-decreasing.

    The rest of your proof seems fine to me (just remove the part where you say "if we choose d so that f is continuous on [x,x+d)." I don't know what that's supposed to mean). What do you think is wrong with it? And what is your new proposition?
     
  8. Oct 6, 2005 #7

    AKG

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    f is continuous from Rl to Rl iff

    [tex](\forall x \in \mathbb{R}_l)(\forall \epsilon > 0)(\exists \delta > 0)(\forall y \in \mathbb{R}_l)\left (y \in [x,\, x+\delta ) \Rightarrow f(y) \in [f(x),\, f(x) + \epsilon)\right )[/tex]

    The problem with your proof is that it does not show that f is non-decreasing. It just shows that for every x, there is an interval to the right of it on which it is non-decreasing. But for x = 0, you might have delta = 1. Then for x = 0.5 you could have delta = 0.5, and for x = 0.99 you could have delta = 0.01. The function could then suddenly drop at 1. Where I said "if we choose d so that f is continuous on [x, x+d)" that was used in conjunction with condition b) so that we know not only that f([x, x+d)) is contained in (f(x) - e, f(x) + e), but that it is contained in [f(x), f(x)+e). We know that [f(x), f(x) + e) is contained in V, but not necessarily that (f(x) - e, f(x) + e) is contained in V. This is why we need for f([x, x+d)) to be contained in [f(x), f(x) + e), and we only get this fact when we say that f is non-decreasing on [x, x+d). I got that it is non-decreasing on that interval by choosing d such that [x, x+d) is continuous, and then used condition b). However, I haven't proven that for every x, there is an interval of the form [x, x+d) on which f is continuous (as a function of R) so that's one reason why the proof is insufficient.
     
  9. Oct 6, 2005 #8

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    Ok, I see the problem now. I was assuming that the following two conditions are equivalent (for any function f:R->R):

    1. for all x, there exists an e>0 such that f([x,x+e)) is a subset of [f(x),∞).
    2. for all a,b in R, b>a implies f(b)≥f(a).

    The former was the one I proved applies here, but the latter would be a more precise definition of "non-decreasing." And, yes, these are not equivalent for general functions, and only coincide when the function is continuous (in the usual sense).

    That being said, your new proposition only slightly modifies the definition of continuity in this space (ie, that the preimage of open sets is open). I think a more interesting assertion would be that a function is continuous in this space iff it is continuous (in the usual sense) and non-decreasing on a set of half open intervals that partition Rl. I haven't tried to prove this yet, and it may be wrong. Another option would be to replace "non-decreasing function" in your original proposition by (1) above.

    By the way, just because I have nothing better to do, here's a proof that the above are equivalent if f is continuous (this may help you if you try to prove my suggestion above):

    It's clear (2) implies (1), since if (2) is true e can be arbitrarily large and (1) will still be satisfied. Assume (1) is true. Now, assume that for some x, there is some n for which f([x,x+n)) is not a subset of [f(x),∞) (This is possible since we have only said there exists an e with the relevant property, not that it is true for any e). Clearly for any e for which f([x,x+e)) is a subset of [f(x),∞), e<n, and so e is bounded above. Since it is a real number, it must have a least upper bound, which we'll call L. Then f([x,x+L)) is a subset of [f(x),∞).

    We also know that there is an e' with the property that f([x+L,x+L+e')) is a subset of [f(x+L),∞). If f(x+L)≥f(x), then e may be taken as L+e', and so L is not the least upper bound of e. So f(x+L)<f(x). Define d by d=f(x)-f(x+L). Then, since f is continuous, the preimage of the interval (f(x+L)-d,f(x+L)+d) is open, and so there is some ε with the property that f((x+L-ε,x+L+ε)) is a subset of (f(x+L)-d,f(x+L)+d). But then for all y in this interval, f(y)<f(x), which contradicts the fact that f([x,x+L)) is a subset of [f(x),∞). Thus there can be no such n, and f([x,∞)) is a subset of [f(x),∞) for all x. Then for any y>x, f(y)≥f(x).
     
  10. Oct 7, 2005 #9

    AKG

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    It certainly need not be continuous in the usual sense. Consider the function which returns the fractional part of the argument. Also, the function does need to be "continuous from the right" in the usual sense. And "non-decreasing on a set of half-open intervals that partition Rl" was sort of what I was trying to get at with the "non-decreasing on intervals where f is continuous."
     
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