Closedness in the proof of the Alaoglu Theorem I'm reading  a proof the Bourbaki-Alaoglu Theorem:

Could someone explain how the closedness of $\Phi(V^\circ)$ (namely, $\Phi(V^\circ)$ contains all of its limit points) is done in the proof?
I totally don't understand the step of  "continuity of the coordinate functions shows that ...". (Where is the continuity used?)
 A: Show that $\Phi(V^o)$ is closed in $K$, which implies by the compactness of $K$ that $\Phi(V^\circ)$ is compact. 
Suppose $\{\Phi(\varphi_\alpha)\}_{\alpha\in A}$ is a net in $\Phi(V^\circ)$ that converges in the product topology to $\psi$ in $K=D^V$. If $x,y,x+y,\lambda x\in V$, then (since by continuity of the coordinate maps, convergence in the product topology is the same as pointwise convergence, which is the same as )
\begin{align*}
\psi(x+y)&=\lim_{\alpha\in A}\left[\Phi(\varphi_\alpha)(x+y)\right]\\
&=\lim_{\alpha\in A}\left[\Phi(\varphi_\alpha)(x)+\Phi(\varphi_\alpha)(y)\right]\\
&=\lim_{\alpha\in A}\Phi(\varphi_\alpha)(x)+\lim_{\alpha\in A}\Phi(\varphi_\alpha)(y)\\
&=\psi(x)+\psi(y)
\end{align*}
and
\begin{align*}
\psi(\lambda x)&=\lim_{\alpha\in A}\left[\Phi(\varphi_\alpha)(\lambda x)\right]\\
&=\lim_{\alpha\in A}\lambda\Phi(\varphi_\alpha)(x)\\
&=\lambda\lim_{\alpha\in A}\Phi(\varphi_\alpha)(x)\\
&=\lambda\psi(x) 
\end{align*}
We have thus shown that $\psi:V\to D$ is linear. Further, $\psi$ determines a map $\hat\psi:X \to\Bbb{F}$ by
$$
\hat\psi(x)=\frac{\psi(t_xx)}{t_x},\quad x\in X
$$
where $t_x>0$ is chosen such that $t_xx\in V$. The existence of $t_x$ is due to the fact that $V$ is absorbing (since it is a neighborhood of $0$).
We have the following observation for $\hat\psi$. Suppose $x\in X$, and $t>0$ such that $tx\in V$. Then we have
$$
\hat\psi(x)=\frac{\psi(tx)}{t}.
$$
Indeed,
$$
\frac{\psi(tx)}{t}=\frac{\psi(\frac{t}{t_x}t_xx)}{t}
=\frac{\frac{t}{t_x}\psi(t_xx)}{t}
= \frac{\psi(t_xx)}{t_x}=\hat\psi(x).
$$
Now we want to show that $\hat\psi\in V^\circ$. First, we show that $\hat\psi$ is linear, i.e., $\hat\psi\in X^*$. Consider $x,y\in X$. Since $V$ is absorbing, there exists $t>0$ such that
$$
tx,ty,t(x+y)\in V.
$$
By the observation above, we have
$$
\hat\psi(x)=\frac{\psi(tx)}{t},\quad 
\hat\psi(y)=\frac{\psi(ty)}{t},\quad
\hat\psi(x+y)=\frac{\psi(t(x+y))}{t}.
$$
It follows that
$$
\hat\psi(x+y)=\frac{\psi(t(x+y))}{t}=\frac{\psi(tx)+\psi(ty)}{t}=\hat\psi(x)
+\hat\psi(y).
$$
Consider $x\in X$ and $\lambda\in \Bbb{F}$. By the fact that $V$ is absorbing again, there exists $t'>0$ such that $tx,t(\lambda x)\in V$ and thus
$$
\hat\psi(\lambda x)
=\frac{\psi(t\lambda x)}{t}
=\frac{\lambda\psi(t x)}{t}
=\lambda\hat\psi(x). 
$$
Now we want to show that $\hat\psi\in V^\circ$. Using the observation above again, we have for every $x\in V$
$$
|\hat\psi(x)|=|\psi(x)|\leq 1.
$$
By definition of $V^\circ$, $\hat\psi\in V^\circ$. Since $\hat\psi|_V=\psi$, we conclude that $\Phi(\hat\psi)=\psi$, which gives the closedness of $\Phi(V^\circ)$.
A: Idea: we have that $V^0$ is homeomorphic (in "right" topologies) to its image $\Phi(V^0)\subset K$ and $K$ is compact. Once we prove that the image is closed then it is also compact, so $V^0$ is a continuous image (under $\Phi^{-1}$) of the compact set, thus, compact.
To prove it is closed they prove that a cluster point of the image belongs to the image, i.e. corresponds to some functional in $V^0$. The details can be like this. Let $k^\alpha$ be a net in the image that converges to $k$. Then there exists a pre-image net $\phi^\alpha$. The functionals are linear, that is $k_{x+y}^\alpha=k_x^\alpha+k_y^\alpha$ etc. The continuity of the coordinate functions means that for any $x\in V$ we have $k_x^\alpha\to k_x$, which gives the same identity for the limits $k_{x+y}=k_x+k_y$. The linearity makes it possible to construct the corresponding linear functional $\phi$ as a pre-image point. Thus $k$ is in the image.
