Does the closure of a smooth algebraic

Let $X\subset \mathbb{C} \mathbb{P}^{N}$ be a smooth, algebraic (locally closed) complex submanifold of $\mathbb{C} \mathbb{P}^N$ of complex dimension $k$. More concretely, $X$ is of the following type $$X := { p \in \mathbb{C} \mathbb{P}^N: \phi_1(p) =0, \phi_2(p) \neq 0 }$$ where $\phi_1$ and $\phi_2$ are sections of some holomorphic vector bundle and whenever $\phi_2(p) \neq 0$, $\phi_1$ is transverse to the zero set.

Let $\overline{X}$ be the closure of $X$ inside $\mathbb{C}\mathbb{P}^N$.

1) Is it true that $\overline{X}-X$ is an algebraic variety?

2) Is it true that the dimension'' of $\overline{X}-X$ is strictly less than the dimension of $X$?

3) In particular does $\overline{X}$ always define a homology class $$[\overline{X}] \in H_{2k}(\mathbb{C} \mathbb{P}^N, \mathbb{Z})$$
the basic idea being that the singularities of $\overline{X}$ are of complex codimension one, hence real codimension two.

Everything is over complex numbers. Note that although $X$ may not be connected, I am assuming that every connected component of $X$ has the same dimension $k$.

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 Sorry, did you mean smooth algebraic (locally closed) submanifold? Then the answer to all the questions is yes. – Serge Lvovski Jan 25 at 16:27 Thank you for the counter example, in my case I am interested only in algebraic (locally closed) sub manifold. Is there a reference for this fact if X is algebaric, ie locally closed? You are saying in that case the answer is yes. – Ritwik Jan 25 at 17:24

1) No. Suppose that $X$ is the set of pairs $(z,w)\in\mathbb C^2$ s.t. $w=e^z$. Then the closure of $X$ in $\mathbb{CP}^2$ is union of $X$ and the line at the infinity (it follows from, say, the big Picard theorem). This is not an algebraic variety.

2) No. In the example above, $\dim(\bar X\setminus X)=\dim X=1$.

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 I have edited the question to indicate what $X$ can be. – Ritwik Jan 25 at 17:30