Prove that there is an $\varepsilon$ such that $f(x) > x + \varepsilon$ for a continuous $f(x) > x$ at $[0,1]$ I know this question was answered by using another theorem here but I wish I could get comments on my way of trying to prove it.
We were asked to prove that for a function $f(x) > x $ which is continuous in $[0,1]$ there exists an $\epsilon$ such that $f(x) > x + \epsilon$ for every $x \in [0,1]$.
I tried a different method then suggested at the link above and would like to know if it's correct and if not what's wrong / missing.
I'll assume the opposite that $\forall \epsilon>0 , f(x) - x < \epsilon$. Also I know that $f(x) - x > 0$ from $f(x) > 0$ so I can deduce that $|f(x)-x| < \epsilon$ that would mean that for any $x \in [0,1]$  the limit of $f$ at that point is $x$, since $f$ is continuous I also know that the limit of $f$ at any $x \in [0,1]$ is $f(x)$ so $f(x)= x$ which is a contradiction to $f(x) > x$ hence there must be at least one $\epsilon$ for which $f(x) - x < \epsilon$
Thanks for any help you can provide
 A: Not sure what you mean by "limit at that point", but you can rework your proof as follows : Suppose the theorem is not true, then for $\epsilon = 1/n, \exists x_n \in [0,1]$ such that
$$
|f(x_n) - x_n| < 1/n \qquad (\ast )
$$
Now $\{x_n\} \subset [0,1]$, which is compact, so there is a convergent subsequence $x_{n_k} \to y \in [0,1]$, and from that and $(\ast )$, you can conclude that $f(y) = y$, which is a contradiction.
In any case, you need to appeal to the compactness of $[0,1]$
A: Well you do need to use the properties of continuous functions on a closed interval. And all the proofs of these properties require the completeness of real number system in a fundamental way. Some proofs above have used the compactness of the closed interval $[0, 1]$ which is again dependent on completeness of real numbers.
Another approach is to the use the following property of continuous functions: if $f(x)$ is continuous on $[a, b]$ then it is bounded on $[a, b]$ i.e. there exist constants $k, K$ such that $k \leq f(x) \leq K$ for all $x \in [a, b]$. A proof can be easily found in any textbook of real analysis or on my blog.
Let $g(x) = f(x) - x$ so that $g(x)$ is continuous on $[0, 1]$ and positive on $[0, 1]$. Then the function $h(x) = 1/g(x)$ is continuous on $[0, 1]$ and hence is bounded on $[0, 1]$. Now let's assume that there is no positive $\epsilon$ which leads to $f(x) > x + \epsilon$ or $g(x) > \epsilon$ for all $x \in [0, 1]$. This means that for any arbitrarily chosen positive number $\epsilon$ we have at least one $x_{\epsilon} \in [0, 1]$ such that $g(x_{\epsilon}) \leq \epsilon$. Then $h(x_{\epsilon}) \geq 1/\epsilon$. Since $\epsilon$ is arbitrary and $1/\epsilon \to \infty$ as $\epsilon \to 0^{+}$, it follows that $h(x)$ is unbounded on $[0, 1]$. This contradiction proves that our assumption is wrong and there exists at least one positive number $\epsilon$ such that $g(x) > \epsilon$ for all $x \in [0, 1]$.
A: The problem is that you are attempting a proof by contradiction, but you're not making your assumptions precise. What do you mean when you say "I'll assume the opposite that $\forall \epsilon>0 , f(x) - x < \epsilon$."? 
You didn't specify, what $x$ refers to. Do you mean $\forall \epsilon \forall x \dots$? Or $\forall \epsilon \exists x \dots$? The latter is the correct negation of the hypothesis, but you seem to be working with the first one.
