proof there are exactly $\mathfrak c$ open sets in $\mathbb R$

Question

There are exactly $\mathfrak c$ open sets in $\mathbb R$

In the proof of the above theorem, there is one line stating that

let $\mathcal I$ be the sets of all open intervals in $\mathbb R$, then $|\mathcal I|=\mathfrak c$.

The proof I am reading does not provide the details for the above statement. I am trying to prove it. Is my argument below right?

Since all the open intervals are in one of the following form: $$(-\infty,a),(a,\infty),or\ (a,b)$$ So the total number of such intervals correspondes to $2$ times the number of ways to choose one number from $\mathbb R$ plus the number of ways to choose two number from $\mathbb R$ Therefore, $$|\mathcal I|=2{\mathfrak c \choose 1}+{\mathfrak c \choose 2}=2\mathfrak c+ \frac {\mathfrak c(\mathfrak c -1)}2 = \mathfrak c$$ This looks very native, is it correct anyway? Could you provide a rigorous way to prove this?

Answer 1

It's clear that there are at least $\mathfrak{c}$ open sets (the intervals $(x,\infty)$, for instance). The set of all open intervals has cardinality $\mathfrak{c}$ as well: the cardinality is $\le\mathfrak{c}+\mathfrak{c}^2+\mathfrak{c}=\mathfrak{c}$.

To finish the proof, note that the rationals are dense in the reals, so every open set is a countable union of open bounded intervals with rational extremes. If $\mathscr{X}$ is the set of all countable families of such open intervals, the map $\mathscr{X}\to\mathscr{U}$ (the topology on $\mathbb{R}$) sending a family to its union is surjective. Therefore $|\mathscr{U}|\le\aleph_0^{\aleph_0}=\mathfrak{c}$.

Answer 2

$(x,x+1)$ is an open interval for each $x$ and this shows that there are atleast $\mathfrak c$ of them. For any open interval $I$ define $f(I)=(a,b)$ if $I$ has finite left end point $a$ and finite right end point $b$, $f(I)=(a,0,0)$ if $I =(-\infty ,a)$ and $f(I)=(0,a,0)$ if $I =(a,\infty )$. Then $f$ is a one-to-one map from the collection of all open intervals to $\mathbb R^{2} \times \mathbb R^{3}$. This last set had cardinality $\mathfrak c$.

Answer 3

Let a rational interval be an interval $(a,b)$ with $a$ and $b$ rational numbers. There are $|\Bbb Q\times \Bbb Q|=\aleph_0\cdot\aleph_0=\aleph_0$ ways to choose these $a$ and $b$, thus there are only $\aleph_0$ rational intervals.

For each open $U\subseteq\Bbb R$, let $V=\{I\subseteq U\mid I\text{ is a rational interval}\}$. The claim is that $\bigcup V= U$. To see this, let $r\in U$, then because $U$ is open, there is some $\varepsilon>0$ such that $(r-\varepsilon,r+\varepsilon)\subseteq U$, and we can find rational numbers $r-\varepsilon<a<r$ and $r<b<r+\varepsilon$ to get a rational interval around $r$ contained in $U$.

So without loss of generality we can consider the number of open sets that are unions of rational intervals. Since we saw there are only $\aleph_0$ rational intervals, there must be $2^{\aleph_0}$ sets of rational intervals, and thus there are at most $2^{\aleph_0}$ unions of rational intervals.

We can easily find $2^{\aleph_0}$ different open sets, by taking the open set $(r,r+1)$ for each $r\in\Bbb R$.

Answer 4

Every open subset of $\mathbb R$ is a union of countably many disjoint intervals. If $\tau$ is the set of all open subsets of $\mathbb R$, then clearly $\,|\tau|\ge |\mathbb R|=c $.

The set $\cal I$ of intervals is equipotent to $\mathbb R$, i.e., $|\cal I|= |\mathbb R|=c$.

Hence, to every open set we can assign a unique countable subsets of $\mathcal I$ (we assume that $\varnothing,\mathbb R\in\cal I$).

If $\mathcal C$ is set of countable subset of $\mathcal I$, then clearly, $|\cal I|\le|\tau|\le |\cal C|$.

Every countable subsets of $\mathcal I$ can be thought of as function $f:\mathbb N\to I$. Clearly, $|\cal I^{\mathbb N}|\ge |\mathcal C|$.

Finally, $$ c=|\mathbb R|\le|\tau|\le |\mathcal C|\le |\mathcal I^{\mathbb N}|=c^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\times\aleph_0}=2^{\aleph_0}=c, $$ and hence $$ |\tau|=c. $$