Analysis continuity proof question Let f: D $\rightarrow$ $\Bbb{R}$  be a continuous at c $\in$ and let $\gamma$ $\in$ $\Bbb{R}$. Suppose f(c) > $\gamma$. Prove there exists $\delta$ > 0 such that
f(x) > $\gamma$ for every x $\in$ B(c;$\delta$) $\cap$ D 
Proof: 
Basically I am a little lost here. I say let $\epsilon$ > 0 and there exists a $\delta$ > 0 such that |f(x) - f(c)| < $\epsilon$ 
So I am just looking at the definition of continunity and find this is almost identical except the ball notation is confusing. What I suggest is 
If c is an isolated point of D then there exist a $\delta$ > 0 such that |x-c| < $\delta$ and x $\in$ D, we have x = c and f(x) = f(c). then f(c) is also > $\gamma$. 
But I am not sure this makes sense completely. I just was trying to interpret what my book is saying. Can anyone give me advice?  
 A: First, by replacing $f$ with $f - \gamma$ we need to show that there's a ball of radius $\delta$ about $c$ such that $x \in B_\delta(c) \implies f(x) > 0$. This gives us one less variable to keep in mind.
The use of balls is a nice notation. Instead of saying $\{x \in D : \lvert x - c \rvert < \delta\}$ we can just say $B_\delta(c)$, the "ball of radius $\delta$ about $c$". Much cleaner and it works not only for the reals but for any $\mathbb R^n$ and later for any metric space. I prefer $B_\delta(c)$ rather than $B(c; ~ \delta)$ but they mean the same thing.
Now whenever you're stuck on what to do, just assume the conclusion is false and see if you can derive a contradiction. That's something I like to do, perhaps that's a matter of taste. I often find that assuming the negation of what I need to prove just gets things going.
So then assume that for every $\delta > 0$ there exists some $x_\delta \in B_\delta(c)$ such that $f(x_\delta) \leq 0$. That's the negation of the conclusion.
Intuitively we would like to extract a sequence from the  $x_\delta$'s, call the sequence $(x_n)$, such that:


*

*$\displaystyle \lim_{n \to \infty} x_n = c$; and

*By continuity of $f$, $\displaystyle \lim_{n \to \infty} f(x_n) = f(c) > 0$; and

*On the other hand, since $f(x_n) \leq 0$ for each $n$, we must have  $\displaystyle \lim_{n \to \infty} f(x_n) \leq 0$.
That's a contradiction. So ... where to find such a nice sequence? Since we can always find an $x_\delta$ for any $\delta > 0$, the natural thing is to let $\delta = \frac{1}{n}$ for $n = 1, 2, 3, \dots$. In other words we have a big disorganized bunch of $\delta$'s and we would like to extract a nice orderly sequence. In these kinds of situations, $\frac{1}{n}$ usually does the trick. You use this idea all the time.
That is, for each positive integer $n$, let $x_n \in B_{\frac{1}{n}}$ with $f(x_n) \leq 0$. Remember that the existence of these $x_n$'s, one for each $B_{\frac{1}{n}}$ is exactly the negation of what we're trying to prove. 
Now we're done and we just have to write out the formal argument from start to finish instead of from finish to start like we just did. So:
Proof: 
We are asked to show that there exists a $\delta > 0$ such that if $x \in B_\delta(c)$ then $f(x) > 0$. 
To prove this by contradiction, assume that instead for every possible $\delta > 0$ there exists some $x_\delta \in B_\delta(c)$ such that $f(x) \leq 0$. I'm subscripting the $x$ to remind myself that each $x$ depends on $\delta$, but if we left out that subscript it would be ok too.
Now for each $n \in \mathbb N^+$ (positive integers) let $x_n \in B_{\frac{1}{n}}$ with $f(x_n) \leq 0$. Such a choice is always possible by our contrary hypothesis. You might want to verify that we negated our quantifiers correctly. That's a frequent point of trouble in these kinds of proofs.
Now we can see that 


*

*$\displaystyle \lim_{n \to \infty} x_n = c$. Why is this? It comes directly from the definition of the limit of a sequence, which you should verify. And ...

*By continuity of $f$, $\displaystyle \lim_{n \to \infty} f(x_n) = f(c) > 0$; and ...

*On the other hand, since $f(x_n) \leq 0$ for each $n$, we must have  $\displaystyle \lim_{n \to \infty} f(x_n) \leq 0$. That's another detail you should prove for yourself.
The second and third bullet points contradict each other, showing that we could NOT make such a choice of $x_n$'s after all. Rather it must be the case that for some $n$, if $x \in B_{\frac{1}{n}} $ then $f(x) > 0$. And then $\delta = \frac{1}{n}$ is the $\delta$ we were required to find. $\square$
By the way note that I never used the definition of continuity. I only used the property of continuous functions that they preserve limits of sequences. The fewer epsilons the better :-) Of course I'm assuming you've already seen that; otherwise it requires proof. And I left out the isolated case but if $c$ is isolated then it has a ball around it with no other points of the domain and $f(c) > 0$ in that case is "all" the points in the ball as you correctly noted.
A: Yes, the claim is clear (as what you've said) if we assumed that $c$ is isolated. 
Suppose $c$ is a cluster point. Take $\epsilon=f(c)-\gamma>0$. Then there exist a $\delta>0$ such that if $x\in D$ and $|x-c|<\delta$ then $$|f(x)-f(c)|<\epsilon.$$ Note that 
$$|x-c|<\delta\iff -\delta<x-c<\delta\iff c-\delta <x<c+\delta\iff x\in (c-\delta,c+\delta).$$ Also,
$$|f(x)-f(c)|<\epsilon\iff -\epsilon<f(x)-f(c)<\epsilon\iff \color\red{f(c)-\epsilon<f(x)}<f(c)+\epsilon.$$ Thus, if $x\in D$ and $|x-c|<\delta$ then
$$f(c)-\epsilon<f(x).$$ This means that if $x\in D\cap(c-\delta,c+\delta)$ then
$$
f(x)>f(c)-\epsilon=f(c)-[f(c)-\gamma]=\gamma.
$$
Take $B(c;\delta)=(c-\delta,c+\delta)$ and we are done.
NOTE: The statement "$x\in D$ and $|x-y|<\delta$" makes sense since $c$ is a cluster point.
