Clarifying a step in the bisection proof of Weierstrass-Bolzano Theorem

Let $(x_n)$ be a sequence of elements of an interval $[a,b]\subset\mathbb{R}$. Then there exists a point of accumulation $c$ of the sequence with $c\in[a,b]$.

The proof goes like this:

Let $I_1=[a,b]$. Let $x_{n_1}\in I_1$. Let $c_1$ be the midpoint of $I_1$. Then $c_1$ separates the interval into $2$ intervals, each of leant $b-a/2$. One of the two intervals must be such that $x_n$ lies in this interval for infinitely many $n$...

How can I translate the bolded part into logical formulas? I think it should be $$(\exists n\in\mathbb{N}_{\geq 1})(\forall m\in\mathbb{N}_{\geq n})(x_m\in[a,c])\lor(\exists n\in\mathbb{N}_{\geq 1})(\forall m\in\mathbb{N}_{\geq n})(x_m\in[c,b]).$$ Now let's say I want to prove the above disjunction by contradiction. The negation is $$(\forall n\in\mathbb{N}_{\geq 1})(\exists m\in\mathbb{N}_{\geq n})(x_m\not\in[a,c])\land(\forall n\in\mathbb{N}_{\geq 1})(\exists m\in\mathbb{N}_{\geq n})(x_m\not\in[c,b]).$$ But I am not able to derive a contradiction from this. Have I translated the bolded statement correctly?

Solution 1:

That is wrong. The assertion$$(\exists n\in\mathbb{N}_{\geq 1})(\forall m\in\mathbb{N}_{\geq n})(x_m\in[a,c])\lor(\exists n\in\mathbb{N}_{\geq 1})(\forall m\in\mathbb{N}_{\geq n})(x_m\in[c,b])$$means that every $x_m$ belongs to $[a,c]$ if $m$ is large enough or every $x_m$ belongs to $[a,c]$ if $m$ is large enough. That is not necessarily true. What is true is that$$(\forall n\in\Bbb N_{\geqslant1})(\exists m\in\Bbb N_{\geqslant n})(x_m\in[a,c])\vee(\forall n\in\Bbb N_{\geqslant1})(\exists m\in\Bbb N_{\geqslant n})(x_m\in[c,b]).$$Denying this means that$$(\exists n'\in\Bbb N_{\geqslant1})(\forall m\in\Bbb N_{\geqslant n'})(x_m\notin[a,c])\wedge(\exists n''\in\Bbb N_{\geqslant1})(\forall m\in\Bbb N_{\geqslant n''})(x_m\notin[c,b]),$$which is false; if such numbers $n'$ and $n''$ existed and if $n=\max\{n',n''\}$, then $x_n\notin[a,c]$ and $x_n\notin[c,b]$. But that's impossible, since $x_n\in[a,b]=[a,c]\cup[c,b]$.