Thread: (Dis)proving that this function is injective

F:ℝ --> ℚ x ℚ x ℚ

In order to be injective every element in my codomain is the image of at most one value of the domain.
It is NOT injective if I find an element in my codomain that is the image of more than one element of the domain.

I think the easiest thing would be to disprove it (Finding a fitting value) but I have a problem of grasping this abstract concept of mapping ℝ on a triple cartesian product.

Similar to how a car mechanic cannot work on your car remotely, until you bring it to his garage, so we can neither prove nor disprove the claim about F until we know what it is.

Well to be honest this is all I have to offer.

My assignment says:
Prove or disprove. There exists an injective image for F:ℝ --> ℚ x ℚ x ℚ

Originally Posted by Cyganek

Well to be honest this is all I have to offer.

My assignment says:
Prove or disprove. There exists an injective image for F:ℝ --> ℚ x ℚ x ℚ

Well $\displaystyle \mathbb{Q} \times \mathbb{Q} \times \mathbb{Q}$ is a countable set.
You may need to prove that.

But $\displaystyle \mathbb{R}$ is uncountable. So what does that tell you?

Ooooooh. A set is countable if it fits the property of being injective.
So if ℝ is NOT countable it means it is NOT injective. Therefore F:ℝ --> ℚ x ℚ x ℚ cant be either.

ℚ consists of the natural numbers and is therefore countable. ℚ x ℚ x ℚ is also countable because the cartesian product of a countable set is ALSO countable, am I right?

I was about to post this.

Actually, I was a bit hasty. If you told your mechanic, "I need my 1900 Honda Accord fixed", he would reply, "There were no Hondas yet in 1900" without needing to see the car. Similarly, any function from $\displaystyle \mathbb{R}$ to $\displaystyle \mathbb{Q}^3$ is not injective. This is because $\displaystyle \mathbb{Q}^3$ is countable, while $\displaystyle \mathbb{R}$ has a strictly greater cardinality of continuum. And if there is an injection from set A to set B, then the cardinality of A is ≤ the cardinality of B.

But it's better to post your question in full.

Thank you very much, it is just so frustrating and enlightening at the same time, thinking about all these problems for hours when I actually KNOW the answer but dont recognize it. For example the cardinality. It makes so much sense and I know this term and what it means, but I somehow didnt put it together with this problem. You guys are awesome. Its impressive how fast you guys answer without any prejudice and so much advice. In German math communitys you get flamed online for not knowing something. Is there any post limit for threads in this forum? I get these assignments weekly and try to solve them on my own, but sometimes I need a hint or two like in this case to get the work done. I have for example 2 other assignments where I am stuck and would like to post them seperately if it is allowed.

Originally Posted by Cyganek

ℚ consists of the natural numbers and is therefore countable. ℚ x ℚ x ℚ is also countable because the cartesian product of a countable set is ALSO countable, am I right?

That is not the reason $\displaystyle \mathbb{Q}$ is countable.

We show that there an injection from $\displaystyle \{r\in \mathbb{Q}:r\ge 0\} \to \mathbb{Z}^+$.
From from which we show $\displaystyle \mathbb{Q}$ is the union of countable sets.

Yea sorry I wrote that a bit too hasty. The first assignment of this kind was actually to proof that Q x Q is countable, where I proved it using a two dimensional chart and how to count Q systematically. I should have been more precise when writing it here, sorry, but yet again thanks for the quick answer.

Originally Posted by Cyganek

A set is countable if it fits the property of being injective.

A set cannot be injective (unless you consider functions in the form of sets). Injectivity is a property of functions, not sets.

Originally Posted by Cyganek

ℚ consists of the natural numbers and is therefore countable.

No, ℚ is the set of rational numbers, but it is still countable.

Originally Posted by Cyganek

ℚ x ℚ x ℚ is also countable because the cartesian product of a countable set is ALSO countable, am I right?

Yes.

Originally Posted by Cyganek

Is there any post limit for threads in this forum? I get these assignments weekly and try to solve them on my own, but sometimes I need a hint or two like in this case to get the work done. I have for example 2 other assignments where I am stuck and would like to post them seperately if it is allowed.

There is no limit. It is recommended to start a new thread for a new problem unless it is closely related to a previous one.