if $f: \mathbb R\to\mathbb R$ is a continuous function , Show that the set $\{x\mid f(x) < p \}$ is open ; proof checking Say that there exists a point $q\in f$ then there must be a $r<p$ such that $f(q)=r$ which is less than $p$, so then there must be a $x$ such that $|f(x)-f(q)|<r<p$. Thus showing by the open definition that it is open.
Is this anywhere near close, or just so far off? And if it is could you give me a direction.
 A: To show that a set is open, it is enough to show that every point in that set is a member of an open interval that is a subset of that set.
The set is $\{x\mid f(x)<p\}$.  Let's call that set $S$.  You say "Say that there exists a point $q\in f$".  Here I take it you mean $q\in S$.  The function $f$ is not the same thing as the set $S$.  So I would say "Suppose $q\in S$.
Then you want to show that $q$ is a member of some open interval that is a subset of $S$.
Since $q\in S$, you know that $f(q)<p$.  You can make use of the fact that $f$ is continuous.
Continuity means for every $\varepsilon>0$, there exists $\delta>0$ such that if $|x-q|<\delta$, then $|f(x)-p|<\varepsilon$.  Notice that the set of points $x$ satisfying $|x-q|<\delta$ is an open interval of which $q$ is a member: it is the interval whose endpoints are $q\pm\delta$.  So there is the question of whether that interval will serve.  In other words, is that interval a subset of $S$?  It is a subset of $S$ if every one of its members $x$ is a member of $S$, and that is true if every such point $x$ satisfies $f(x)<p$.  We know that every such point $x$ satisfies $|f(x)-f(q)|<\varepsilon$.  Does that imply $f(x)<p$?  The statement $|f(x)-f(q)|<\varepsilon$ is the same as saying $f(x)$ is between the two numbers $f(q)\pm\varepsilon$.  If $f(x)<f(q)+\varepsilon$, does that imply $f(x)<p$?
Look at that last question:
If $f(x)<f(q)+\varepsilon$, does that imply $f(x)<p$?
Let's put it this way: What would $\varepsilon$ need to be in order for that to work?
A: Hint: 
Our goal is to show that for any $x$ such that $f(x)<p$, there exists some $\delta>0$ such that for any $y$ such that $|x-y| < \delta$, $f(y) < p$.
Why is this the same as saying "the set $\{x\mid f(x) < p \}$ is open"? How can I use the continuity of $f$ to show that this is the case?
