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Let $T = \{\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7}\}$ and $\mathcal{S}$ the set of all functions that maps $T$ into $\mathbb{Q}.$ What is the cardinality of $\mathcal{S}$?

So far I've been messing with the idea that we can take the subset $T_0\subset T$ of functions that square an element of $T$ and adds any other string of rational numbers to it. For example $f(\sqrt{2})=(\sqrt{2})^2+a$ where $a \in \mathbb{Q}.$ Since $|T_0|$ is countable and infinite $|T_0|=\aleph_0.$

I'm not really sure if this is going to get me anywhere.

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    $\begingroup$ It is same as the cardinality of $\mathbb{Q}^{4}$, which is countable. $\endgroup$ – Seewoo Lee Nov 24 '18 at 5:15
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    $\begingroup$ The particular values of your set are not important. Only that it has four elements. $\endgroup$ – badjohn Nov 24 '18 at 6:27
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Could you answer the question of the cardinality of the set of all functions from $\{ 1 \}$ to $\mathbb{Q}$? I hope so, it's obviously the same as $\mathbb{Q}$ since any such function can be specified by a single value from $\mathbb{Q}$.

Now, how about the cardinality of the set of all functions from $\{ 1, 2 \}$ to $\mathbb{Q}$? Well, any such function can be specified by two values from $\mathbb{Q}$ so its cardinality is the same as $\mathbb{Q} \times \mathbb{Q} = \mathbb{Q}^2$. Do you know the cardinality of this set?

Let's jump a step to the set of functions from $\{1, 2, 3, 4 \}$ to $\mathbb{Q}$. As Seewoo says, this has cardinality $\mathbb{Q} \times \mathbb{Q} \times \mathbb{Q} \times \mathbb{Q} = \mathbb{Q}^4$. If you figured that $\mathbb{Q}^2$ is countable then you should see that this is as well.

Now your set: $\{\sqrt{2},\sqrt{3}, \sqrt{5}, \sqrt{7}\}$. Well there is a very simple bijection between it and my set $\{1, 2, 3, 4 \}$ which can be easily used to associate the maps from my set to $\mathbb{Q}$ with those from your set to $\mathbb{Q}$.

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  • $\begingroup$ I absolutely understand this. Thank you so much. I got so caught up in the values of the set and not how many elements are in the set. This is a new level of abstraction for me compared to what I've been exposed to so far at the undergraduate level. $\endgroup$ – Trevor Mason Nov 25 '18 at 0:36
  • $\begingroup$ @TrevorMason. I expect that was the intention of whoever wrote the question. How about the set $\{ 0, 1, e, i, \pi \}$? $\endgroup$ – badjohn Nov 25 '18 at 8:42
  • $\begingroup$ This should have cardinality $\mathbb{Q}^5?$ $\endgroup$ – Trevor Mason Nov 25 '18 at 21:39
  • $\begingroup$ Yes, but what is that? $\endgroup$ – badjohn Nov 25 '18 at 22:12
  • $\begingroup$ $\mathbb{Q}= \aleph_0?$ $\endgroup$ – Trevor Mason Nov 25 '18 at 22:13

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