If $f $ is right continuous at $c $, can I claim that $|sup _{x \in [c,x _i ]} f(x) -inf_{x \in [c,x _i ]}f(x)|<\epsilon $? I realized I needed to make a quite big edit...
The question should be if it is possible to show that $\forall \epsilon > 0 \ \exists x_i: |sup _{x \in [c,x _i ]} f(x) -inf_{x \in [c,x _i ]}f(x)|<\epsilon $?
And not if $f $ is right continuous at $c $, can I claim that $\forall \epsilon > 0 \ \exists x_i: |f(c)-sup _{x \in [c,x _i ]} f(x) |<\epsilon  $?
Well if a function $f $ is continuous on a closed and bounded interval, it is uniformely continuous on this interval and I'm wondering if some similar statement can be made about a right or left continuous function?
Thanks in advance!
 A: Yes, to both questions. In essence, you are restricting the domain of the function to the part which is continuous. To get your first claim, we have that for every $\epsilon > 0$, there exists some $x_i > c$ such that $|f(c) - f(x)|< \epsilon$ whenever $x \in [c,x_i]$, so certainly $\sup_{x \in [c,x_i]} |f(c) - f(x)| < \epsilon$. Now note that $|f(c) - \sup_{x \in [c, x_i]} f(x)| \leq \sup_{x \in [c,x_i]} |f(c) - f(x)|$.
This is a response to "So to answer ...", but I cannot respond to it because of reputation restrictions. Note that the $x_i$ may be different for the sup and int. It may be more appropriate to use $x_1$ for the sup and $x_2$ for the inf. Let $x_i$ be the minimum of $x_1$ and $x_2$ and you have your conclusion - just as you argued below.
A: So to answer my edited question, I start from where J. Im showed that $|f(c)-sup _{x \in [c,x _i]}f(x)|\le \epsilon $. Then as $|f(c)-f(x)|< \epsilon $ when $x \in [c,x _i ]$ we have that $|f(c)-inf _{x \in [c,x _i]}f(x)|< \epsilon $.
And from the triangle inequality $|sup_{x \in [c,x _i]}f(x)-inf_{x \in [c,x _i]}f(x) |\le 2 \epsilon $.
Correct?
