How does one prove the convergence of a trajectory in the Ramsey model? Consider the Ramsey-Cass-Koopmans model in discrete time. Specifically, consider the following phase diagram.



For simplicity, let's consider the trajectory marked A which is in the northwest corner of this diagram. 

In this location, supposedly $K_{t+1}&lt;K_t$ and $C_{t+1}&gt;C_t$. But how does one show this? Up to this point, all I have seen is that "supposedly" this trajectory veers off, but I don't see how it necessarily does so. 

Let $$C_t = K_t^\alpha + (1-\delta)K_t - K_{t+1}$$

My Quesiton

Can someone explain why A veers off towards the NW direction? Specifically, why must $K_{t+1}&lt;K_t$ and $C_{t+1}&gt;C_t$?
See Romer (2006, Sec 2.3) for details of notation.
Dynamics are given by
\begin{align}
\dot k &amp;= f(k) - (n+g)k - c\\
\dot c &amp;= \frac{c}{\theta}(f'(k) - \rho - \theta g)
\end{align}

Presume you are located somewhere on the $\dot k = 0$ locus. Now if we increase $c$ one gets
\begin{align}
\frac{\partial \dot k}{\partial c} = -1 &lt; 0
\end{align}

That is, if we are above $\dot k = 0$ capital decreaes over time. Hence at point $A$ we move to the left.

Now presume you are located somewhere on the $\dot c = 0$ locus. Now if we increase $k$ one gets
\begin{align}
\frac{\partial \dot c}{\partial k} = \frac{c}{\theta} f''(k) &lt; 0
\end{align}

That is, if we are right of $\dot c = 0$ consumption decreaes over time. Hence at point $A$ we move up in addition (left of $\dot c = 0$).