Proof using contradiction and a contrary assumption

Question

I've been given this problem for multi-variable calculus, which isn't too difficult to prove using the definition of a limit;

Let $f:\Bbb{R}^n\to\Bbb{R}$ be a continuous function and let $\mathbf x_0\in\Bbb{R}^n$. Assume that $f(\mathbf x_0)>0$ and prove that there are positive $a>0$ and $r>0$ such that $f(\mathbf x)>a, \forall\mathbf x\in B(\mathbf x_0, r)$

However, we've been challenged to set up a proof by contradiction, and more specifically to first formulate a 'contrary assumption'. Although, I'm unsure what that is exactly and was wondering if someone could help. I believe it means setting up a logical negative as follows;

Assume that $f(\mathbf x_0)>0$ and that there exists no positive $a>0$ and $r>0$ such that $f(\mathbf x)<a, \exists\mathbf x\notin B(\mathbf x_0, r)$

but I'm having trouble using that, which makes me think it's incorrect.

Cheers

Answer 1

We are asked to prove:

if $f(x_0) > 0$, then there are $a>0$ and $r>0$ such that $f(x)>a,∀x∈B(x_0,r)$.

Its negation will be:

$f(x_0) > 0$ and not: there are $a>0$ and $r>0$ such that $f(x)>a,∀x∈B(x_0,r)$.

In "logical form" it is:

$\lnot [ \exists a>0 \ \exists r>0 \ ∀x∈B(x_0,r) (f(x)>a)$].

Moving inside the negation sign, we have:

$\forall a>0 \ \forall r>0 \ \exists x∈B(x_0,r) (f(x) \le a)$.

Thus, the requested contradictory assumption will be:

$f(x_0) > 0$ and for every $a>0$ and every $r>0$ there is an $x∈B(x_0,r)$ such that $f(x) \le a$.

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To check it, we can consider its meaning; the original sentence amounts to saying that if $f(x_0)$ is positive, we can always find a positive $a$ and a "suitable" interval around $x_0$ such that $f(x)$ is greater than $a$ inside it.

Thus, its negation must be: $f(x_0)$ is positive but for $a$ positive whatever, in every interval around $x_0$ there is always a point where the function $f(x)$ is smaller than $a$.