Proof that a set is uncountable and also equipotent with the real numbers.

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I'm looking at some old exam questions to prepare and I have problems with one particular question. I struggle with cardinality.

The question goes as follows:

We call $B \subset \mathbb{R}$ an open set if:

$\forall x \in B:\exists\alpha>0:\forall y\in \mathbb{R}:|x-y|<\alpha\Rightarrow y\in B.$

a) Show that $ a,b \in \mathbb{R}$ with $a<b$ exist so that $]a,b[ \subset B$.

b) proof that $B$ is uncountable.

c) proof that $B$ is equipotent with $\mathbb{R}$.

For a) I don't see a way to do this in a good way. I have some problems with fully understanding this question.

For b) and c) I guess I can proof that the cardinality is neither finite nor equal to $\aleph_0$. Should I next find 2 injective functions so that with the Schröder–Bernstein theorem I can say that they are equipotent?

I know you better show what you already have but starting these problems seems the hardest for me so I don't really have much.

Thank you in advance.

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This is false if $B=\emptyset$. I will assume otherwise.

a) Take $x\in B$. Then there is a $\alpha>0$ such that $(x-\alpha,x+\alpha)\subset B$. So, take $a=x-\frac\alpha2$ and $b=x+\frac\alpha2$.

b) You know that $B\supset(a,b)$ for some $a,b\in\mathbb R$. Since $(a,b)$ is uncountable, then so is $B$.

c) Since $B\subset\mathbb R$, $\lvert B\rvert\leqslant\lvert\mathbb R\rvert$. But $B\supset(a,b)$, which has the same cardinal as $\mathbb R$. So, $\lvert B\rvert=\lvert\mathbb R\rvert$.