Posted by: holdenlee | May 25, 2010

B7s: May 15, 2010


  1. “Last night’s temperature took all values between -3 and 5.” Show that it suffices to say “Both -3 and 5 occured among last night’s minimum temperatures (by location).” Assume that temperature is a continuous function of time and location.
  2. Let H\subseteq \mathbb{R}, H\neq \{0\} be closed under addition. Let f:H\to\mathbb{R} be nonincreasing and additive (f(x+y)=f(x)+f(y);x,y\in H). Show that f(x)=cx on H where c\geq 0.
  3. Let A,B be (additive) abelian groups. For \nu, \chi homomorphisms from A to B, define their sum by (\nu+\chi)(a)=\nu(a)+\chi(a). Then the set of homomorphisms from A to B forms an abelian group. Suppose A is a p-group (p prime).
    1. Prove that H becomes a topological group under the topology by taking p^kH,k\in \mathbb{N} to be a neighborhood base around 0.
    2. Prove that H is complete in this topology.
    3. Prove that every connected component consists of a single element.
    4. When is H compact?
  4. For what x\in \mathbb{C} does there exist a proper subfield \mathbb{F}\subset \mathbb{C} such that \mathbb{F}(x)=\mathbb{C}?

Solutions (Sketches)

  1. Use intermediate value theorem twice, once for time and once for location.
  2. Proof follows the same argument as in \mathbb{R} (Cauchy’s Equation).
    1. Show that negation is a continuous function H\to H and addition is a continuous function H\times H\to H. The first is obvious. For the second, note that if \alpha+\beta=\gamma then (\alpha +p^kH)+(\beta+p^kH)\subseteq \gamma+p^kH((\alpha +p^kH),(\beta+p^kH)) is an open set in H\times H.
    2. Take the metric defined by d(\nu,\chi)=p^{-l} where l= -\max \{k \mid \nu - \chi \in p^kH\}. Given a Cauchy sequence \{\chi_i\}, we define \chi as follows: for a with order p^k by the Cauchy condition \chi_i will eventually stay in the same coset \alpha+p^kH. Then \chi_i(a) stabilizes; set \chi(a) equal to this value. One can show that \chi is a homomorphism and \chi_i\to \chi.
    3. It suffices to show that each open set is disconnected; it suffices to show that p^kH can be written as a union of disjoint closed sets. Just write p^kH=p^{k+1}H\cup (p^kH-p^{k+1}H). (the complements of either can be written as a union of cosets=open sets in the form \alpha+p^iH.)
    4. If [p^kH:p^{k+1}H] is finite for each k, then the proof of compactness is similar to the proof in \mathbb{R}^n. Suppose there’s an open cover without a finite subcover; by induction and Pigeonhole Principle take a coset \alpha_k+p^kH without a finite subcover. Take \chi_k\in \alpha_k+p^kH; this Cauchy sequence converges to some \chi by completeness. There must be an open set in the cover containing \chi, but it contains \alpha_k+p^kH for large enough k, contradiction.
      Else [H:p^kH]=\infty for some k. Then the cosets of p^kH form an open cover without a finite subcover.
      This reduces the problem to a group theory problem…
  3. Working on it.


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