Prove that for $x\in \mathbb{R}$ $\exists n\in \mathbb{N}$ such that $n>x$

real-analysissolution-verification

I solved this and I think it is correct but being new to Real Analysis without much proofs experience there is always this hesitancy I have during/after solving so I am posting this in 'solution-verification'

Prove that for $x\in \mathbb{R}$ $\exists n\in \mathbb{N}$ such that
$n>x$

My attempt:

Assume that this statement is incorrect, this means that for $x\in \mathbb{R}$ $\nexists n\in \mathbb{N}$ such that $n>x$

So we now have that $\forall x \in \mathbb{R}$ $x\geq n$ $\forall n \in \mathbb{N}$

Since we are talking about all $x \in \mathbb{R}$ and $\mathbb{N}\subset\mathbb{R}$ we can say that $\forall x \in \mathbb{N}$ $x\geq n$ $\forall n \in \mathbb{N}$

This implies that $\mathbb{N}$ is bounded above which is wrong, thus our initial statement must be correct.

Any feedback or help would be appreciated. My goal from this is to either verify it is correct or get to know where I went wrong so I am not really interested in someone giving me another solution.

Best Answer

The given statement is

$\forall x \in \mathbb{R} \,\exists n \in \mathbb{N}$ such that $n >x$

If the above statement is incorrect then it should be

$\exists x \in \mathbb{R} \,\forall n \in \mathbb{N}$ such that $n \le x$

and not

So we now have that $\forall x \in \mathbb{R}$ $x\geq n$ $\forall n \in \mathbb{N}$

Edit: Your can use the same argument. Remember the definition of bounded subsets of real numbers. A subset $A$ of real numbers is said to be bounded if $\exists M \in \mathbb{R} \forall a \in A$ we have $a \le M$.

Here you can take $A= \mathbb{N}, M = x$.

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