Is the set of finite rank operators with rank at most n closed w.r.t the weak operator topology? I know the set of finite rank operators with rank less than n is closed in strong operator topology. Can we say it is also closed in weak operator topology?
 A: EDIT: There is a very short proof in case you have the result for one of the mentioned operator topologies available: SOT (Strong Operator Topology) and WOT (Weak Operator Topology) generate the same topological dual space and thus convex sets are SOT-closed iff they are WOT-closed. Note that linear subspaces of vector spaces are convex and thus the desired result follows immediately since the set of all operators of rank at most $k$ is a linear subspace.

Now to the original proof I provided:
For all $k \in \mathbb{N}$, let $ \mathfrak F_k(X,Y) = \mathfrak F_k$ denote the set of finite rank operators with rank at most $k$. 
Showing that $\mathfrak F_k$ is closed w.r.t the strong operator topology (SOT) is similar to showing that $\mathfrak F_k$ is closed w.r.t the weak operator topology (WOT) since on normed spaces of finite dimension norm convergence and weak convergence are equivalent.
We start by proving the following 
Lemma. Let $(w_1^i)_{i \in I}, \dots, (w_k^i)_{i \in I} \subseteq Y$, $k \in \mathbb{N}$, be nets converging weakly to the linearly independent vectors $w_1,\dots,w_k$. Then there exists $i_0 \in I$ such that for all $i \geq i_0$ the vectors $w_1^i, \dots, w_k^i$ are linearly independent.
Proof. Consider the linear subspace $V:= \langle w_1, \dots, w_k\rangle$ and the  corresponding dual basis $w_1^*, \dots, w_k^*$. Since, $V$ is finite dimensional $w_1^*, \dots, w_k^*$ can be extended continuously to functionals on $Y$ (Hahn-Banach) which we will again denote by $w_1^*,\dots,w_k^*$.
We can now explicitly write down the continuous projection $p_V$ onto $V$:
$$
p_V(x) = w_1^*(x) w_1 + \dots + w_k^*(x) w_k.
$$
As $w_1^i,\dots,w_k^i$ converge weakly, their images under $p_V$ will converge strongly. You can check out e.g. this post to see that the vectors $p_V(w_1^i),\dots, p_V(w_k^i)$ will eventually become linearly independent and so will their preimages $w_1^i,\dots,w_k^i$.$\quad\square$
Now, let $(f_i)_{i \in I} \subseteq \mathfrak{F}_k$ be a SOT-convergent net. Suppose $f := \operatorname{WOT-}\lim_i f_i \not \in \mathfrak{F}_k$. Then, there exist linearly independent vectors $\eta_1, \dots, \eta_{k+1} \in \operatorname{range} f$ with corresponding linearly independent preimages $\xi_1, \dots, \xi_{k+1}$, i.e.
$$
f(\xi_{1}) = \eta_1,\dots, f(\xi_{k+1}) = \eta_{k+1}.
$$
Since $(f_i(\xi_j))_{i \in I}, j = 1,\dots,{k+1},$ converge weakly, by the above lemma there exists $i_0 \in I$ such that $f_{i_0}(\xi_1),\dots,f_{i_0}(\xi_{k+1})$ are linearly independent. This contradicts the assumption that all $f_i$ were of at most rank $k$. Thus, we have proved that
$$
\mathfrak F_k = \overline{\mathfrak F_k}^{\operatorname{WOT}} \left(= \overline{\mathfrak F_k}^{\operatorname{SOT}} \right)
$$
for all $k \in \mathbb{N}$.
