In a separable Hilbert space, how to show that the orthogonal projection onto a subspace of $n$ orthonormal basis elements converge? Could anyone help me with this problem? I don't know where to start.

Let $\{ e_n \}_{n=1}^\infty$ be an orthonormal basis in a separable Hilbert space $H$. Denote by $P_n$ the orthogonal projection onto the subspace spanned by $e_1, \dots ,e_n$.

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*Show that $P_nx \rightarrow x$  as $n \rightarrow \infty$ for any $x \in H$ and, furthermore, the convergence is uniform on any compact subset of $H$.

*Show that for any compact operator $T$ on $H$ we have $\Vert P_nT - T \Vert \rightarrow 0$ as $n \rightarrow \infty$. Therefore $T$ is approximated by the finite rank operators $P_nT$.


Thanks for any help!
 A: *

*Let $x\in X$. We have $P_n(x)=\sum_{j=1}^n\langle x,e_j\rangle e_j$ so , by Bessel-Parseval equality 
$$\lVert P_nx-x\rVert^2=\sum_{j\geq n+1}|\langle x,e_j\rangle|^2.$$
As the latest series is convergent we have the result. 
Now, let $K\subset H$ compact. Fix $\varepsilon >0$. Then we can find an integer $N$ and  $x_1,\dots,x_N\in K$ such that for each $x\in K$, we can find $1\leq k \leq N$ such that $\lVert x-x_k\rVert\leq\varepsilon$. Fix $x\in K$. Then $$\lVert P_nx-x\rVert^2=\sum_{j\geq n+1}|\langle x-x_k+x_k,e_j\rangle|^2\leq \varepsilon^2+\max_{1\leq k\leq N}\sum_{j\geq n+1}|\langle x_k,e_j\rangle|^2.$$
As the RHS doesn't depend on $x$, we have
$$\sup_{x\in K}\lVert P_nx-x\rVert^2\leq \varepsilon^2+\max_{1\leq k\leq N}\sum_{j\geq n+1}|\langle x_k,e_j\rangle|^2.$$
Now take the $\limsup_{n\to+\infty}$ to get the result.

*If $T$ is compact, then $K:=\overline{T(B(0,1))}$ is compact, so we apply the previous result to this $K$.


Note that the property of approximation of a compact operator by a finite rank operator is true in any Hilbert space, not only in separable ones. To see that, fix $\varepsilon>0$; then take $v_1,\dots,v_N$ such that $T(B(0,1))\subset \bigcup_{j=1}^NB(y_j,\varepsilon)$. Let $P$ the projection over the vector space generated by $\{y_1,\dots,y_N\}$ (it's a closed subspace). Consider $PT$: it's a finite ranked operator. Now take $x\in B(0,1)$. Then 
pick $j$ such that $\lVert Tx-y_j\rVert\leq\varepsilon$. We also have, as $\lVert P\rVert\leq 1$, that $\lVert PTx-Py_j\rVert\leq \varepsilon$. As $Py_j=y_j$, we get $\lVert PTx-Tx\rVert\leq 2\varepsilon$.
