Let $X$ be a normed space, $Y$ a dense subspace of $X$ and $Z$ a closed finite-codimensional subspace of $X$. Is $Z\cap Y $dense in $Z$ ?
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Suppose first that the codimension of $Z$ is $1$. Then $Z=\ker\phi$ for some nonzero $\phi\in X^*$. There is some $y\in Y$ such that $\phi(y)\neq0$. Define $P:X\to X$ by $$Px=x-\frac{\phi(x)}{\phi(y)}y.$$ Then $P$ is linear and bounded, $PX=Z$, and $PY\subset Y\cap Z$. If now $z\in Z$, there is a sequence $\{y_n\}$ in $Y$ converging to $x$. Thus $\{Py_n\}$ is a sequence in $Y\cap Z$ converging to $Pz=z$.
In general, if $Z$ has codimension $n$, then $Z=\bigcap_k\ker\phi_k$ for some nonzero $\phi_1,\ldots,\phi_n\in X^*$, and we can find $y_1,\ldots, y_n\in Y$ such that $\phi_k(y_k)\neq0$. Then define $P:X\to X$ by $$Px=x-\sum_{k=1}^n\frac{\phi_k(x)}{\phi_k(y_k)}y_k.$$
yes, it is dense in $Z$. draw a picture with one disk and a secant line, you will find the secret. the line is a finite dimension space, there are many points belong to $Y$ around the line but not in the line, because the line-finite dimension space is nowhere dense(suppose $X$ is infinite dimensional)