For $f,g~(f0$ let $\{h\in\mathcal C[0,1]:t-c
For $f,g~(f<g),t\in\mathcal C[0,1],c>0$ let $\{h\in\mathcal C[0,1]:t-c<h<t+c\}$$=\{h\in\mathcal C[0,1]:f<h<g\}.$ I want to show that $t-c=f,~t+c=g.$
$$t-c<t<t+c\text{ and } \\f<\dfrac{f+g}{2}<g.\\\text{Then }t-c<\dfrac{f+g}{2}<t+c\text{ and } f<t<g.$$
I don't know how to contradict the following cases:
For some $y\in[0,1].$

*

*Let $t(y)-c>f(y)$


*Let $t(y)-c<f(y)$


*Let $t(y)+c>g(y)$


*Let $t(y)+c<g(y)$

This problem can more clearly be written as:
For $f_1,g_1,f_2,g_2\in\mathcal C[0,1]$ $$\{h\in\mathcal C[0,1]:f_1<h<g_1\}=\{h\in\mathcal C[0,1]:f_2<h<g_2\}\implies f_1=f_2,g_1=g_2.$$


Let $A=\{h\in\mathcal C[0,1]|\;t-c<h<t+c\}$ and $B=\{h\in\mathcal C[0,1]|\;f<h<g\}$.
For every $\epsilon\in(0,c)$ we have $t-\epsilon,t+\epsilon\in A$ by definition of $A$. Since $A=B$, this means that $t-\epsilon,t+\epsilon\in B$ for all $\epsilon\in(0,c)$, which by definition of $B$ means that $$f<t-\epsilon<t+\epsilon<g.$$ This implies that $$f\leq t-c<t+c\leq g.$$ To complete the proof, we have to show that $f\geq t-c$ and $g\leq t+c$. To do this, note that (since $f<g$) for every $\epsilon\in(0,1)$ we have $$f<(1-\epsilon)f+\epsilon g<g,$$ therefore $(1-\epsilon)f+\epsilon g\in B$. But this means that $(1-\epsilon)f+\epsilon g\in A$ for all $\epsilon\in(0,1)$, which means that $$t-c<(1-\epsilon)f+\epsilon g<t+c.$$ Now, let $\epsilon\to 0$. This gives us $$t-c\leq f\leq t+c.$$ Similarly, letting $\epsilon\to 1$ gives us $$t-c\leq g\leq t+c.$$ This completes the proof.
