I'm trying to solve a three part exercise.
The first part asks me to show that the set $T=\{(x,y)\in I\times I : x < y\}$, where $I$ is just some open interval in $\mathbb{R}$ is connected.
I've done this by showing that it's path connected (just take a straight line path between any two points)
The second part - the one I'm having trouble with - asks me to show that, for a differentiable function $f:I\rightarrow\mathbb{R}$ and for $g: T \rightarrow \mathbb{R}$ defined by
$$g(x,y) = \frac{f(x) - f(y)}{x-y}$$
that we have $g(T)\subseteq f'(I) \subseteq\overline{g(T)}$, where $\overline{g(T)}$ is just the closure of $g(T)$.
I can see how to solve the third part which comes after this provided I can do this part, but I'm really struggling to see how to show these inclusions.
Any help you could offer would be really appreciated.
The first inclusion, as stated in the comments, is a consequence of Lagrange's theorem. As for $f'(I) \subseteq \overline{g(T)}$, a possible argument would be the following. Any point in $\overline{g(T)}$ is the limit of a $g(T)$-valued sequence. Now, consider some $z \in f'(I)$, so $z = f'(x)$ for some $x \in I$. This implies that the following equality holds (in particular, the following limit exists), \begin{equation} \lim_{\varepsilon \rightarrow 0}\frac{f(x+\varepsilon)-f(x)}{\varepsilon} = f'(x). \end{equation} This implies that the sequential limit exists, and in particular that the sequential left limit exists, so that for some $I$-valued sequence $x_n$ converging to $x$ from the left (i.e. for every $n$, $x_n < x$) we have \begin{equation} \lim_{n \rightarrow +\infty}\frac{f(x_n)-f(x)}{x_n - x} = f'(x). \end{equation} Now, for any $n$ we have that $(x_n, x) \in T$ since $x_n < x$, so $(x_n, x)$ is a $T$-valued sequence. Call it $r_n := (x_n, x)$, and thus the equation turns into \begin{equation} \lim_{n \rightarrow +\infty} g(r_n) = f'(x), \end{equation} which shows that $z$ is the limit value for some $g(T)$-valued sequence and thus belongs to its closure.