<center> <img src = "https://i.imgsafe.org/be/be7ef20ee7.png"/></center>
[1]

# Infinitely many tasks in Finite amount of time?

Today I will post about old but gold philosopical paradoxes; all are considered as __supertasks__ . 

## 0. What is a Supertask?
A __supertaks__ is a task that consists in infinitely many component steps, but which in some sense is completed in a finite amount of time. Supertasks were studied by the pre-Socratics and continue to be objects of interst to moder philosophers. The term "supertask" itself was first used by [Jame F.Thomson](https://en.wikipedia.org/wiki/James_F._Thomson_(philosopher)) in 1954. Among many examples of supertasks, I will post about supertasks related to __mechanics__ since we can model as well as analyze these tasks using the concept in physics. 

## 1. Zeno's paradox

<center><img src = "https://i.imgsafe.org/c8/c88a203492.jpeg"/></center>
[2]

The most famous example is the Zeno's paradox. The runner begins at <img src="http://latex.codecogs.com/gif.latex?x=0" title="x=0" /> and runs toward the finish line at <img src="http://latex.codecogs.com/gif.latex?x=1" title="x=1" />. 

- First he runs <img src="http://latex.codecogs.com/gif.latex?1/2" title="1/2" align = "center"/> of the distance, so that the current position is at <img src="http://latex.codecogs.com/gif.latex?x=1/2" title="x=1/2" align = 'center'/> . 

- Next he runs half the remaining distance, so that the current position is at <img src="http://latex.codecogs.com/gif.latex?x=\frac{1}{2}&space;&plus;&space;\frac{1}{4}" title="x=\frac{1}{2} + \frac{1}{4}" align = "center" /> . 

- Again he runs half of the remaining distance, so that the current position is at <img src="http://latex.codecogs.com/gif.latex?x=\frac{1}{2}&space;&plus;&space;\frac{1}{4}&space;&plus;&space;\frac{1}{8}" title="x=\frac{1}{2} + \frac{1}{4} + \frac{1}{8}" align = "center"/> . 

- Formally, let's define the step <img src="http://latex.codecogs.com/gif.latex?T_n\&space;(n&space;\geq&space;2)" title="T_n\ (n \geq 2)" align = "center"/> to be the task performed by runner running from 

<center><img src="http://latex.codecogs.com/gif.latex?x&space;=&space;\frac{1}{2}&space;&plus;&space;\frac{1}{2^2}&space;&plus;...&plus;&space;\frac{1}{2^{n-1}}" title="x = \frac{1}{2} + \frac{1}{2^2} +...+ \frac{1}{2^{n-1}}" /> </center> 

to 

<center> <img src="http://latex.codecogs.com/gif.latex?x&space;=&space;\frac{1}{2}&space;&plus;&space;\frac{1}{2^2}&space;&plus;...&plus;&space;\frac{1}{2^{n-1}}&space;&plus;&space;\frac{1}{2^{n}}" title="x = \frac{1}{2} + \frac{1}{2^2} +...+ \frac{1}{2^{n-1}} + \frac{1}{2^{n}}" /> </center>

. He continues in this way infinitely many times, getting closer to the finish line. Now Zeno's question was 

<center> <b><i> Can he ever complete these infinite number of tasks? </i></b> </center>

---

Zeno answered this question as NO. We all know that Zeno was wrong, but why did he answered as no? 

The question from Zeno, needs to be clarified since the meaning of __completion__ is vague. To solve this paradox, we need to clarify the word __complete__ . 

1. First, "complete" can refer to the execution of a final action (step). This sense of completion does not occur, since for every step in the task (say <img src="http://latex.codecogs.com/gif.latex?T_n" title="T_n" align = "center"/>) there is another step that happens later, (for example, <img src="http://latex.codecogs.com/gif.latex?T_{n&plus;1}" title="T_{n+1}" align = "center"/>) . 

2. On the other hand, "complete" can refer to __carrying out every step__ in the task, which certainly does occur because of the convergence of geometric series 

<center> <img src="http://latex.codecogs.com/gif.latex?\sum_{n=1}^{\infty}&space;\frac{1}{2^n}&space;=&space;\frac{1/2}{1-1/2}&space;=&space;1" title="\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1/2}{1-1/2} = 1" /> </center>

The two meanings for the word "complete" happen to be equivalent only for __finite taks__ , where most of our intuitions about tasks are developed. But they are not equivalent when it comes to supertasks! Unfortunately, Zeno did not realize the difference between finity and infinity. His __misunderstanding__ of the concept of infinity, lead to paradoxical situation that the runner can not reach the finish line. 

---

To sum up, the runner can carry out every step, but there is no  __final__ step. Is this the ultimate answer to all supertask paradoxes? Well, let's look at another example. 

## 2. Thomson's Lamp

In 1954, J.F Thomson came up with an ingenius example of supertask, called Thomson's Lamp.

<center> <img src = 'https://i.imgsafe.org/c8/c86efa2a52.jpeg'/></center>
[3]

Suppowe we are given a lamp. 

- At time <img src="http://latex.codecogs.com/gif.latex?t=0" title="t=0" /> we switch off a lamp. 

- After 1 minute, at time <img src="http://latex.codecogs.com/gif.latex?t=1" title="t=1" align = "center"/> we switch it on. 

- After <img src="http://latex.codecogs.com/gif.latex?1/2" title="1/2" align = "center"/> minute more, at time <img src="http://latex.codecogs.com/gif.latex?t=1&plus;1/2&space;=&space;\frac{3}{2}" title="t=1+1/2 = \frac{3}{2}" align = 'center'/> we switch it off again. 

- After <img src="http://latex.codecogs.com/gif.latex?1/4" title="1/4" align = "center"/> minute more, at time <img src="http://latex.codecogs.com/gif.latex?t&space;=&space;1&space;&plus;&space;1/2&space;&plus;&space;1/4&space;=&space;\frac{7}{4}" title="t = 1 + 1/2 + 1/4 = \frac{7}{4}" align = "center"/> we switch it on again. 

- Formally, the lamp is off at time interval 

<center> <img src="http://latex.codecogs.com/gif.latex?\left[&space;1&space;&plus;&space;\frac{1}{2}&space;&plus;&space;...&plus;\frac{1}{2^n},&space;1&space;&plus;&space;\frac{1}{2}&space;&plus;&space;...&plus;\frac{1}{2^{n&plus;1}}\right&space;)" title="\left[ 1 + \frac{1}{2} + ...+\frac{1}{2^n}, 1 + \frac{1}{2} + ...+\frac{1}{2^{n+1}}\right )" /> </center>

if <img src="http://latex.codecogs.com/gif.latex?n" title="n" /> is odd, and on if <img src="http://latex.codecogs.com/gif.latex?n" title="n" /> is even. 

The difference between the Zeno run and Thomson's lamp is that the Zeno run is continuous (the runner's position is continuous on the real line), while state of the lamp is discrete; either on or off. Summing each of the time intervals above gives rise to an infinite geometric series that converges to 2 minutes, 

<center> <img src="http://latex.codecogs.com/gif.latex?\sum_{n=0}^{\infty}&space;\frac{1}{2^n}&space;=&space;\frac{1}{1-1/2}&space;=&space;2" title="\sum_{n=0}^{\infty} \frac{1}{2^n} = \frac{1}{1-1/2} = 2" /> </center>

after which time the entire supertask has been completed. Now the question is 

<center> <b><i> When 2 minutes is up, is the lamp on or off? </i></b> </center> 

---

If you read section 1, you must realize that the question we asked above is same as asking the final state of the lamp. So using the conclusion of section 1, such state can not exist. Is this true? A lamp that is neither on nor off after 2 minutes...&#128533; 

One might argue that such lamp that switches  the state infinitely many times can not exist in real world. But this is not a mathematical argument, since switching states infinitely many times is totally valid in theoretical sense. 

---

To solve this paradox, we must clarify the situation. Define a function <img src="http://latex.codecogs.com/gif.latex?f" title="f" align = "center"/> such that 

<center> <img src="http://latex.codecogs.com/gif.latex?f(t)&space;=&space;\begin{cases}1&space;&&space;(\text{if&space;lamp&space;is&space;on&space;at&space;time&space;}t)\\&space;0&space;&&space;(\text{if&space;lamp&space;is&space;off&space;at&space;time&space;}t)&space;\end{cases}" title="f(t) = \begin{cases}1 & (\text{if lamp is on at time }t)\\ 0 & (\text{if lamp is off at time }t) \end{cases}" /> </center>

Then, <img src="http://latex.codecogs.com/gif.latex?f(t)&space;=&space;0" title="f(t) = 0" align = "center"/> for all values in interval <img src="http://latex.codecogs.com/gif.latex?[0,&space;1)" title="[0, 1)" align = "center"/> , <img src="http://latex.codecogs.com/gif.latex?f&space;=&space;1" title="f = 1" align = "center"/> for all values in  <img src="http://latex.codecogs.com/gif.latex?[1,&space;3/2)" title="[1, 3/2)" align  = "center"/> and so on. For convenience, denote the interval 

<center> <img src="http://latex.codecogs.com/gif.latex?\left[&space;1&space;&plus;&space;\frac{1}{2}&space;&plus;&space;...&plus;\frac{1}{2^n},&space;1&space;&plus;&space;\frac{1}{2}&space;&plus;&space;...&plus;\frac{1}{2^{n&plus;1}}\right&space;)" title="\left[ 1 + \frac{1}{2} + ...+\frac{1}{2^n}, 1 + \frac{1}{2} + ...+\frac{1}{2^{n+1}}\right )" /> </center>

as <img src="http://latex.codecogs.com/gif.latex?I_n" title="I_n" align = "center"/> . Now our question is equal to asking <img src="http://latex.codecogs.com/gif.latex?f(2)" title="f(2)" align = "center"/> given that the function <img src="http://latex.codecogs.com/gif.latex?f" title="f" align = "center"/> is defined on 

<center> <img src="http://latex.codecogs.com/gif.latex?[0,&space;1)&space;\cup&space;\left(&space;\bigcup_{n=1}^{\infty}&space;I_n&space;\right&space;)" title="[0, 1) \cup \left( \bigcup_{n=1}^{\infty} I_n \right )" /> </center>

Hmm... is 

<center> <img src="http://latex.codecogs.com/gif.latex?2&space;\in&space;[0,&space;1)&space;\cup&space;\left(&space;\bigcup_{n=1}^{\infty}&space;I_n&space;\right&space;)&space; ?" title="2 \in [0, 1) \cup \left( \bigcup_{n=1}^{\infty} I_n \right )" /> </center>

If does, there should exist an interval <img src="http://latex.codecogs.com/gif.latex?I_n" title="I_n" align = "center"/> such that <img src="http://latex.codecogs.com/gif.latex?2&space;\in&space;I_n" title="2 \in I_n" align = "center"/>. But this is nonsense since 

<center> <img src="http://latex.codecogs.com/gif.latex?1&plus;&space;\frac{1}{2}&space;&plus;&space;...&space;&plus;&space;\frac{1}{2^{n&plus;1}}&space;<&space;2" title="1+ \frac{1}{2} + ... + \frac{1}{2^{n+1}} < 2" /> </center>

for all values of <img src="http://latex.codecogs.com/gif.latex?n" title="n" /> . So, what we are trying to obtain is the function value __outside__ of the domain! Thus, our answer to this question has __no right answer__ . 

<center> <b> <i> It can be either on or off after 2 minutes </i></b></center> 

which is independent of the lamp's switching task on <img src="http://latex.codecogs.com/gif.latex?t&space;\in&space;[0,&space;2)" title="t \in [0, 2)" align = "center" /> .  The description of the Thomson lamp only actually specifies what the lamp is doing at each finite stage before 2 minutes. It says nothing about what happens at 2 minutes, especially given the lack of a converging limit. It may still be possible to “complete” the description of Thomson’s lamp in a way that leads it to be either on after 2 minutes or off after 2 minutes. The price is that the final state will not be reached from the previous states by a convergent sequence. 

## 3. Example of Lamp  turned on at t = 2

Suppose a metal ball bounces on a conductive plate, bouncing a little lower each time until it comes to a rest on the plate. Suppose the bounces follow the same geometric pattern as before. Namely, the ball is in the air for 1 minute after the first bounce, <img src="http://latex.codecogs.com/gif.latex?1/2" title="\frac{1}{2}" align = "center"/> minute after the second bounce, <img src="http://latex.codecogs.com/gif.latex?1/4" title="\frac{1}{2}" align = "center"/> minute after the third, <img src="http://latex.codecogs.com/gif.latex?1/8" title="\frac{1}{2}" align = "center"/> minute after the fourth, and so on. Then the entire infinite sequence of bounces is a supertask. 

Now suppose that the ball completes a circuit when it strikes the metal plate, thereby switching on a lamp. This is a physical system that implements Thomson’s lamp. In particular, the lamp is switched on and off infinitely many times over the course of a finite duration of 2 minutes.

<center> <img src = "https://i.imgsafe.org/c8/c86ecf1eaa.jpeg" /></center>
[4]

## 4. Example of Lamp turned off at t = 2

Alternatively, we could arrange the ball so as to break the circuit when it makes contact with the plate. This gives rise to another implementation of Thomson’s lamp, but one that is off after 2 minutes when the ball comes to its final resting state.


<center> <img src = "https://i.imgsafe.org/c8/c86ed11635.jpeg" /></center>
[5]

You can see that if the ball touches the metal plate, current does not flow through the wire passing through the lamp. 

## 5. What is the difference?

The difference between the  two supertasks is due to the construction of function <img src="http://latex.codecogs.com/gif.latex?f" title="f" align = "center"/> . Since we already constructed a function for Thomson's Lamp, let's do the same for Zeno's paradox. 

Define the runner's  velocity function as <img src="http://latex.codecogs.com/gif.latex?v(t)" title="v(t)" align = "center"/> . It is reasonable to assume that <img src="http://latex.codecogs.com/gif.latex?v" title="v" /> is continuous over <img src="http://latex.codecogs.com/gif.latex?[0,&space;1)" title="[0, 1)" align = "center"/>, so that the position function is given by 

<center> <img src="http://latex.codecogs.com/gif.latex?f(t)&space;=&space;\int_0^{t}&space;v(x)dx" title="f(t) = \int_0^{t} v(x)dx" /> </center>

on <img src="http://latex.codecogs.com/gif.latex?t&space;\in&space;[0,&space;1)" title="t \in [0, 1)" align = "center" /> using the [fundamental theorem of calculus](https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part). Again, fundamental theorem of calculus assures the __uniform continuity__ of <img src="http://latex.codecogs.com/gif.latex?f" title="f" align = "center"/> on <img src="http://latex.codecogs.com/gif.latex?[0,&space;1)" title="[0, 1)" align = "center"/> . Now, this guarantees the existence of __left-hand limit__

<center> <img src="http://latex.codecogs.com/gif.latex?\lim_{t&space;\rightarrow&space;1-}f(t)" title="\lim_{t \rightarrow 1-}f(t)" /> </center>

which is equal to 

<center><img src="http://latex.codecogs.com/gif.latex?\sum_{n=1}^{\infty}&space;\frac{1}{2^n}&space;=&space;1" title="\sum_{n=1}^{\infty} \frac{1}{2^n} = 1" align = "center" /></center>

using the Zeno's paradox. The [most surprising fact](https://math.stackexchange.com/questions/245237/extension-of-a-uniformly-continuous-function-between-metric-spaces) is that the extension of <img src="http://latex.codecogs.com/gif.latex?f" title="f" align = "center"/> by defining 

<center> <img src="http://latex.codecogs.com/gif.latex?f(1)&space;=&space;\lim_{x&space;\rightarrow&space;1-}f(x)&space;=&space;\sum_{n=0}^{\infty}&space;\frac{1}{2^n}&space;=&space;1" title="f(1) = \lim_{x \rightarrow 1-}f(x) = \sum_{n=0}^{\infty} \frac{1}{2^n} = 1" /> </center>

is indeed __unique continuous extension__ of <img src="http://latex.codecogs.com/gif.latex?f" title="f" align = "center"/> ! Therefore, the runner's position after carrying out all the steps should be <img src="http://latex.codecogs.com/gif.latex?x=1" title="x=1" />; no other choice available. 

## 6. Conclusion

- Zeno's paradox can be modeled as a continuous function over half open interval, and this has unique continuous extension which gives unique consequence after completion of all the steps.

- However, Thomson's lamp paradox can be modeled as a discontinuous (furthermore, discrete) function over half open interval, and this has multiple extensions (both 0 and 1 are possible outcomes). 

-  Look we started from simple paradox, and ended up with real analysis...&#128517;
## 7. Citations

[1] Selfmade by GeoGebra 5.0

[2] [Image Source Link](https://plato.stanford.edu/archives/win2016/entries/spacetime-supertasks/#MissFinaInitStepZenoWalk)

[3] Self Made using Drawing App

[4] Self Made using Drawing App

[5] Self Made using Drawing App

---

<center> <img src = "https://imgoat.com/uploads/3278a4a869/112096.PNG"/></center>
<center> <img src = "https://steemitimages.com/DQmU7Xv1NnXmSE1MAYNN1eM6R5Xs2Cdtczp9DpaopmzopGE/rocket.gif"/></center>