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I feel as if this question is closely related to Topology, but I have not taken any Topology, so I am currently having trouble proving this for a Set Theory course. (Unsure if I should tag this under topology as well, so correct me if I am wrong in doing so.)

Question: Let S be the set of open subsets of the real. Show that S is in 1-1 correspondence with $\mathbb{N^N}$.

From venturing across the internet I found out what an open subset is and what this set S will sort of look like.

All I have done is:

Pf. Let $S \subseteq \mathbb{R} $, and let $x \in S$.

If x is rational, we can define $ I_x= \cup_{x \in I} I $

If x is irrational, for $ \epsilon > 0, (x - \epsilon , x+ \epsilon) \subseteq U $.

Then we have a rational $y$ such that $ y \in (x - \epsilon , x+ \epsilon) \subseteq U $

Thus $ x \in I_y$ and $\forall x \in U, x \in I_{\alpha}$ for some $ \alpha \in U \cap \mathbb{Q} $

Then we have that $ U= \cup_{\alpha} I_{\alpha}$.

How do I use this to show a 1-1 with $\mathbb{N^N}$? Or am I going about it all wrong? Any suggestions, hints, corrections are appreciated.

Once again, I am taking a shot at the dark at this, please be kind.

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  • $\begingroup$ First find out the cardinality of the set of open subsets of the reals and then show it is the same as N^N . $\endgroup$ – Jacob Wakem Nov 4 '16 at 6:58
  • $\begingroup$ What set theory axioms are you using? This is important. $\endgroup$ – Jacob Wakem Nov 4 '16 at 6:59
  • $\begingroup$ At this point, none. This is still on the naive side of set theory. I haven't delved into the axiomatic aspect of set theory. $\endgroup$ – El Spiffy Nov 4 '16 at 7:00
  • $\begingroup$ N^N has the same cardinality 2^N is the cardinality of the reals. Thus we wish to show that there are only as many open subsets of the reals as there are real numbers (which seems reasonable considering open sets are very special compared to sets) $\endgroup$ – Jacob Wakem Nov 4 '16 at 7:18
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You can read the following topic: 'Every open set in $\mathbb{R}$ is the union of disjoint open intervals.' How do you prove this without indexing intervals with $\mathbb{Q}$?

There, they talk about the fact that each open set of Reals is actually union of disjoint open intervals(which are countably many). Now, fix some arbitrary ordering $\omega$ of these countably many intervals, as well as some arbitrary ordering $\tau$ of the set of rationals $\mathcal{Q}$.

Now, we can define our map $\phi$. The value of $\phi$ for some open set $\mathcal{O}$ is the sequence $a(\omega , \tau) = (a_{1},a_{2},\ldots )$, such that:

$a_i = 0$, if $\tau (i)\notin \mathcal{O}$

and $a_i = k$, where $\omega (k)$ is the interval from the representation of $\mathcal{O}$, that contains $\tau (i)$.

This map is close to bijective, since for any two different open sets(disjoint unions of open intervals) there are either a rational number that belongs to only one of the sets or if they have the same elements, then we must have two rationals that are in the same interval for one of the open sets and intervals with different numbers for the other set.

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