Tag Archives: leftovers

Turducken Day 6, also, answer to Cantor set problem

12 Jun

Honestly we didn’t have that many leftovers: I packed up a bunch with my friends who came over to eat the turducken.

The best thing I made with the leftovers was turducken tacos, mostly because the fatty meat was a wonderful complement to the chipotle yogurt sauce inspired by Mark Bittman’s fish taco recipe.

Turducken tacos:

Leftover turducken, in heated flour tortillas, with lettuce, chopped radishes, cheddar cheese, and chipotle yogurt sauce.  Serve with a wedge of lime.



2 c full-fat plain yogurt

2 minced garlic cloves

3 chopped up chipotle peppers in adobo sauce

Leave in a fridge for at least ten minutes: it just gets better the longer it sits.

Also, I made matzoh ball soup with the turducken broth leftover.  It was also quite tasty, though salty (must dilute the broth)

Don't DILLy-dally, eat it while it's hot!

Don’t DILLy-dally, eat it while it’s hot!

Finally, the answer to the Cantor set question from that other post:

This answer is straight from an exercise in Bruckner, Bruckner, and Thomson‘s book Real Analysis (exercise 1.1.3).

First, think about the number 0.5637.  In grade school we say this is five-tenths plus six-one-hundredths plus three one-thousandths plus seven ten-thousandths.  We can write this as a sum as $\latex \frac{5}{10}+\frac{6}{10^{-2}}+\frac{3}{10^{-4}}+\frac{7}{10^{-5}}$.  Switching to fancy math notation, we can write any number as a sum: \sum_{n=1}^{\infty} \frac{a_n}{10^{-n}}.  Here, \sum_{n=1}^k means that you take whatever’s after the sigma symbol, and start at n=1 and add.  So in our example, we have 0.5637 = \sum_{k=1}^4 a_k 10^{-k}, where $a_1=5, a_2=6, a_3=3, a_4=7$.

This decimal expansion is relevant.  I said earlier that the Cantor set has something to do with ternary expansions.  So in our decimal expansions, we let $a_k = 0,1, \ldots, 9$.  For a ternary expansion, we can only let $a_k = 0,1,2$.  And for our Cantor set, it’s all the numbers in [0,1] that don’t include the digit 1 in their ternary expansion.

Let’s use another set to get to our answer.  Let D = \{ x\in [0,1]: x=\sum_1^{\infty} \frac{j_n}{3^n}, j_n=0,1\}, that is, D is the points in the interval [0,1] with no 2 in their ternary expansion.

Pick any point y\in [0,1] and its ternary expansion \sum_1^{\infty} \frac{y_n}{3^n}.  We’re going to find numbers a,b\in D so that a+b=y, which will prove that D+D=[0,1].  If y_n = 2, let a_n=b_n=1.  If y_n=1, let a_n=1, b_n=0.  If y_n=0, let a_n=b_n=0.  This way we’ve defined sequences a_n,b_n, and if we let a= \sum_1^{\infty}\frac{a_n}{3^n}, b=\sum_1^{\infty}\frac{b_n}{3^n}, we’ve defined a,b\in D so that a+b=y, as desired.

This doesn’t quite finish our problem!  We wanted to show that C+C = [0,2].  But notice that $D=\frac{C}{2}$, that is, if some number x is in C, x/2 is in D, and if y is in D, then 2y is in C.  So if D+D=[0,1], then C+C = [0,2], and we’re done!

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