here are afew comments/hints:4. SInce $P$ in non-zero there is a vector $v$ so t

here are afew comments/hints:4. SInce $P$ in non-zero there is a vector $v$ so that $w=Pv ne 0$.Since $P$ is a projection it maps $w$ to itself. Thus the norm of $P$is at least $1$.7. Suppose $M $ is an ideal not of this form. Then for every point $x$in $S$ there is a function $f in M$ that does not vanish at $x$.Thus $|f|^2 = f cdot bar f$ is in the ideal is non-negative and isstrictly positive on a open neighborhood of $x$. Using compactnesscover $S$ by a finite number of such neighborhoods and add theresulting functions; we get an element $g$ of $M$ that is bounded awayfrom zero. Thus $1 = g cdot (1/g)$ is in $M$ so $M = C(S)$ acontradiction.9. In each of these we use the fact that in a metric space compactness isthe same a sequential compactness. For example if $x_n +y_n$ is asequence in $C+C$ then just choose a subsequence where $x_n$converges and then a further subsequence where $y_n$ converges.Then $x_n + y_n$ converges. The other parts are similar.10. If an operator $T $ is invertible then its inverse is bounded and hencethere is a positive $c>0$ so that $| Tx | geq c |x|$ for every vector$x$. But the proof of Theorem 6 in Chapter 21 shows that for a compactoperator this can happen only on a finite dimensional subspace.11. Consider the operators $M_n$ on the sequence space $ell^2$ ofsquare summable sequences that are the identity of the first $n$ coordinatesand set the rest to zero (i.e. these are the orthogonal projections ontothe first $n$ coordinates). They all have finite dimensional range henceare compact but converge strongly to the identity which is not compact.
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