James Tauber's Blog 2004/11/01
Too Taxing for the Servers
Online submissions of Australian federal income tax returns were due today. (Normally they are due 31st October, but given it was a Sunday, they made it Monday 1st November).
I completed my return last night and went to lodge it and it came back with an Error 1200 and told me to call the e-Tax technical hotline. By this stage it was about 2am so I decided to wait until it was working hours on the east coast (give them a chance to come in to work and see that something was broken).
So early this morning, I tried lodging again. Error 1200. So I called the hotline number. Engaged. Lodge again. Error 1200. Hotline still giving busy signal.
After a few hours, I stopped getting a busy signal and got put on hold (listening to piano works on what sounds like a radio slightly tuned off the station). While on hold, I tried lodging a few more times. Still getting an Error 1200.
Finally got through to someone. I decided I wouldn't be mad at the guy—he's probably received hundreds of "Error 1200" calls already today. I didn't even need to tell him the issue - he just asked right away if I was getting an Error 1200.
Turned out some servers had been down since 5pm Sunday. He told me they should be back up by this evening but that the Australian Tax Office had extended lodgement until Wednesday for e-Tax filers.
So I just tried lodging my tax at 11.55pm this evening. No Error 1200. But it told me I'd already lodged!
by jtauber : Created on Nov. 1, 2004 : Last modified Feb. 8, 2005 : (permalink)
Poincare Project: Open Sets
In Open Balls and Continuity, I said:
Imagine that you don't know the distance function of either the domain or co-domain of the function but someone who does has precalculated all the open balls for you. Of course, for most metric spaces, there would be an infinite number of these, but the key point here is that you only need to know what the open balls are to test continuity. You don't need to know the distance function.
So imagine that a friend has given you a set along with all of the subsets that are open balls. From this alone, you can establish whether a function on the set is continuous.
We can simplify the definition of continuity (and other concepts) by introducing a more general subset than the open ball called an open set. Even though we will initially define open sets in terms of open balls, we can simply provide the open sets without reference to open balls, much like your friend provided the open balls without reference to the metric.
A subset of a set X is called an open set if it is the union of open balls of X.
Now continuity can be defined in terms of open sets (and this definition can be proven to be the same as that using open balls). A function from X to Y is continuous if and only if, for each open set in Y, the inverse in X is also an open set.
So, instead of giving you the open balls, your friend could give the open sets. And from that, you'd be able to establish whether a function was continuous.
Now you might be wondering: in the absence of the original metric, how would we know whether the collection of open sets was really a collection of open sets and not just some random selection of subsets? (Perhaps you don't trust your friend.) Well, it turns out open sets have some interesting properties:
- the union of any collection of open sets in X is also an open set in X;
- the intersection of any finite collection of open sets in X is also an open set in X;
- the empty set is open;
- the set X itself is open.
What is so significant about this is that a collection of subsets of X is a collection of open subsets of X if (and only if) it has the four properties above. It doesn't matter if it came from a distance function or open balls or some random selection of subsets. As long as the four properties above hold, the subsets are open subsets and can be used to demonstrate continuity (along with many other things).
NOTE: It is important that the union can be of any collection whereas the intersection can only be of a finite collection.
For a while, I thought the second two rules might be redundant and derivable from the first but a number of people (including Michael Walter and Richard Plagge) have clarified it for me. Michael points out the case where X = {1, 2} and the candidate collection of sets is {{1}}. This meets the first two rules but not the second two. Richard gives the example X = {a, b, c, d} with the candidate collection {{a}, {a, b}, {a, b, c}}. Again, the first two rules are met but the second two clearly do not follow.
UPDATE: next post
by jtauber : Created on Nov. 1, 2004 : Last modified Aug. 10, 2007 : Categories poincare_project : 0 comments (permalink)