Blowing up $\mathbb{A}^{2}$ in the origin I've some problems with the construction of the blow-up of $\mathbb{A}^{2}$ in the origin

The set-up/construction:

Put $X = \mathbb{A}^2$ with coordinates denoted by $x_{0},x_{1}$ and let $f_{0} = x_{0}, f_{1} = x_{1}$. I now want to construct the blow up of $X$ in $(f_{0}, f_{1} )$.
First I consider the graph:
$\Gamma = \lbrace ((p_{0},p_{1}),(p_{0}:p_{1})) \vert (p_{0},p_{1}) \in U \rbrace$ where $U = \mathbb{A}^2/(0,0)$. The blow up of $X$ is then defined to be the closure of $\Gamma$ in $\mathbb{A^{2}}\times \mathbb{P}^1$ (we let the coordinates in $\mathbb{P^{1}}$ be denoted by $(y_{0}:y_{1})$). In order to find this closure I use the fact that $\Gamma$ can be covered by $D(x_{1})\times D(y_{1})$ and $D(x_{0})\times D(y_{0})$.
(Here I use the notation $D(x_{i} ) = \lbrace (x_{0}, x_{1}) \vert x_{i} \neq 0 \rbrace$  and $D(y_{i}) = \lbrace (y_{0}:y_{1}) \vert y_{i} \neq 0\rbrace $ for $i = 1,2$.)
On $D(x_{0})\times D(y_{0} )$ we can write $\Gamma = Z(x_{0}\frac{y_{1}}{y_{0}}-x_{1})$ and similarly we get on $D(x_{1})\times D(y_{1})$ that $\Gamma$ can be written as $\Gamma = Z(x_{1}\frac{y_{0}}{y_{1}}-x_{0})$.
Because of this we conclude that every point $((x_{0},x_{1}),(y_{0}:y_{1}))$ in $\Gamma$ must satisfy $x_{0}y_{1} = x_{1}y_{0}$ and thus we have that the closure (the blow up of $X$, denoted by $\tilde{X}$) is given by the zero locus of $x_{0}y_{1} - x_{1}y_{0}$ in $\mathbb{A}^2\times\mathbb{P}^1$.


Question:
I understand why $\Gamma \subset \tilde{X}$ and that $\tilde{X}$ is closed. But why is $\tilde{X}$ the smallest set containing $\Gamma$? This detail bothers me. Maybe a dimension argument would work? I hope someone can clarify this.
 A: Consider a line $\ell$ through the origin in $\mathbb{A}^2$, and the corresponding thing $\ell'$ in $\Gamma$.  Since $\ell' \subset \Gamma$, we have $\operatorname{cl}(\ell') \subseteq \operatorname{cl}(\Gamma)$.  We can check that


*

*$\ell'$ is not already closed, but

*$\ell'$ together with one additional point is closed.


Therefore, $\operatorname{cl}(\ell')$ consists of $\ell'$ together with this one additional point.  As such, this one additional point lies in $\operatorname{cl}(\Gamma)$.  Moreover, each point of $\widetilde{X} \setminus X$ comes from such a line, so $\widetilde{X}$ is at least contained in $\operatorname{cl}(\Gamma)$.  You've already shown that it contains $\operatorname{cl}(\Gamma)$ above, so you're done.
The idea of the blowup is that we replace the origin, which is a single point, with one point for every possible slope going through the origin.  A curve $C$ passing through the origin of $X$ with tangent $m$ becomes a curve passing through the point $(0, m)$ of $\widetilde{X}$.  In particular, this means that if $C$ passes through the origin more than once, but with different tangents, e.g. if it looks like this:

then the proper transform $\widetilde{C}$ will not self-intersect at the origin.
