Extract 6.6
Context: Extracts 6.6i and ii come from the two individual sessions with students Alan and Connie, students of the same tutor. The reason I present them together is that they both refer to CD2.5 and illustrate the two novices' different but characteristic difficulties with the notions of sup and inf. In Alan's session the tutor is demonstrating what he calls a 'more attractive' way to do CD2.3&4 and an alternative-to-Alan's way to CD2.1ii. The latter is also how Connie's tutorial starts. In this it turns out that Connie needs to be reminded of the Archimedean Property, as well as be given a pictorial representation of the proof. After CD2.1ii, in both sessions the students inquire about CD2.5, which is where the Extracts presented here start.
The Episode: Extract 6.6i
The tutor says he does not approve of Alan's proof (I have reconstructed Alan's proof in fig.6b). He then explains that Alan is interpreting a statement for a particular x in SÈT (xÎSÈT means xÎS or xÎT, stresses the tutor) as a statement for all xŒS. It is, the tutor says, similar to saying that because f(x)³g(x) for some x in the domain of f and g, hence f³g.
The tutor and Alan then agree that CD2.5ii is not true. Alan cannot recall the counterexample he used. The tutor offers a counterexample (S={1,2}, T={1,3}) and asks Alan to think where exactly he realised that CD2.5ii was not true.
A1: It's true that supS is an upper bound for the intesection but it's just saying that... it breaks down when I prove that it is the least upper bound... because if there is an element in between...
T1: It would look awfully like this again!
A2: Oh, it's the same... probably... so supS is 2...
T2: Hmm... so alpha could go one and a half from there...[Alan nods] It's the problem about that you are considering two alternatives for all x as opposed to it holding for a particular x... I'm not certain...
A3: I'm saying... that x here... I think it's this... it's quite difficult to say...
The tutor says there is a 'shakier' counterexample which requires a bit of 'trickery': choosing S and T such that SÇT=Æ then SÇT has no supremum. He however suggested the one given above because he thought of it as 'slightly more convincing'.
The Episode: Extract 6.6ii
Connie says she has problems with CD2.5i so, at her request, the tutor presents CD2.5i (see fig.6a). To have a supremum the union must be non-empty and bounded above. To prove the latter the tutor says he is 'forking out' the cases for xÎSÈ T: if xÎS then x£supS. Similarly if xÎT then x£supT. Tutor is then interrupted by Connie who points at his writing (reconstructed from the recording in fig.6a) and says about x:
C1: Well, one of them... x is in the bit where they...
T1: No, no x is somewhere in the union.
C2: Well, then isn't it in both of them?
T2. The tutor gives the definition of SÈT, as the set of all elements in S and T, and, returning to his proof, continues his sentence: that in either case, whether xÎS or xÎT, x£ supS or supT. Connie then interupts again:
C3: So you don't need to consider both?
T3: No, I don't need to. If either of this works then it's enough for me. So in either case x£max of the two sups.
C4: That's it?
No, he replies, this only shows that SÈT is bounded above and, since it is non-empty, that it has a supremum. The tutor completes the presentation (see fig.6a) while Connie is nodding. For part ii she says that 'On a second thought I think I had to find a counterexample'. Then they go on with CD2.6.
Return to Appendices for Chapter 6.