Every partition $P$ $\delta$-fine has $c$ as tag if and only if $\delta(x)\leq |x-c|$ Let $\delta$ a gauge of $[a,b]$. How can I prove the following?
Every $\delta$-fine tagged partition $\mathcal{P}$ of $[a,b]$ has $c$ as tag $\Leftrightarrow$ $\delta(x)\leq|x-c|$ for all $x\in[a,b]-\{c\}$.
I'm try to link the things I know about Gauges and partitions to solve this, but I'm not getting it.
I know that $\delta(x)\leq|x-c| \Leftrightarrow c\in[x-\delta(x),x+\delta(x)]\,\forall x\in[a,b]-\{c\}$. 
If $\mathcal{P}=\left\{\left(I_{i},t_{i}\right)\right\}_{i=1}^{n}$ is a $\delta$-fine partition, $I_{i}\subset\left[t_{i}-\delta(t_{i}),t_{i}+\delta(t_{i})\right]$. I have to prove that $t_{j}=c$ for some $j$
 A: I believe that the only change that must be made on the hypotheses is that on the right side of the equivalence, $\delta (x) < \lvert x - c \rvert$; i.e., the inequality is strict. The proof is by contradiction.
If we assume  that $\delta (x) < \lvert x - c \rvert$, then suppose that you have a tagged partition $\dot P$, $c \in I_j=[p_j,p_{j+1}]$, where $I_j$ is an interval of your partition, and suppose that $t_j \neq c$ is the tag of $I_j$ (without loss of generality, assume $t_j > c$). Then (draw a figure!), $$t_j -\delta (t_j) > t_j - (t_j-c)=c$$ But $c\geq p_j$. So $$t_j - \delta (t_j)>p_j$$ which contradicts that $\dot P <<\delta$, since we should have $p_j \in [t_j - \delta(t_j),t_j + \delta(t_j)]$. For the converse, the contradiction is that if we didn't have $\delta(x)<\lvert x-c \rvert$, then we could construct a partition that is $\delta$-fine, but that doesn't have $c$ as a tag. This is fairly easy to do: Suppose that for some $x \in [a,b]$, $\delta(x)\geq \lvert x-c \rvert$. Consider the partition which has the interval $I_j$ with lower endpoint $p_j = c$, tag $t_j = x$ and upper endpoint $x+\delta(x)$. Then, this partition is $\delta$-fine and doesn't have $c$ as a tag, which is the contradiction we needed. $\square$
