Determine whether S is a subspace of $P_n$, the vectorspace of all real polynomial Determine whether S is a subspace of $P_n$, the vector space of all real polynomials of degree $\leq n-1 $ of this form:
\begin{equation}
p = a_0 + a_1X+a_2X^2+ \dots + a_{n-1}X^{n-1}
\end{equation}
To determine if the set $S = \left\{ p \in p_n(\mathbb{R}) \mid\forall \alpha \in \mathbb{R}: p(\alpha) = p(-\alpha) \right\}$ is a subspace, I need to check for these 3 things:


*

*let q, p $\in S$, then $r = q+p \in S$

*let $\mathbb{B} \in \mathbb{R}$ and $p \in S$, then $\mathbb{B}p \in S$

*$\mathbf0 \in S$


So for number 1, we let $r(a) = q(a) + p(a)$ and because $q(a) + p(a) = q(-a) + p(-a)$ then $r(a) = r(-a)$. 
And for number 2, we let $w(a) = \mathbb{B}p(a)$. and because $\mathbb{B}$ it's just a scaler $w(-a) = \mathbb{B}p(-a)$.
But how do I check for number 3, i.e showing that $\mathbf0 \in S$?
 A: Other people have clarified to you what you are missing.  I am presenting something that may be helpful to you in the future.  After you learn about linear independence or linear maps, you can come back here and read this answer again.

If you already have some knowledge about linear maps, then you can show that $S$ is an $\mathbb{R}$-vector subspace of $P_n$ of dimension $\left\lceil\dfrac{n}{2}\right\rceil$ by exhibiting a surjective linear map $\varphi:P_n\to\mathbb{R}^{\left\lfloor\frac{n}{2}\right\rfloor}$ and showing that $S=\ker(\varphi)$.  Since the kernel of a linear map from a vector space $V$ to another vector space $W$ is a vector subspace, the claim follows (using the Rank-Nullity Theorem).
Now, I shall give you a good map $\varphi$.  Define
$$\varphi(p):=\big(p(+1)-p(-1),p(+2)-p(-2),p(+3)-p(-3),\ldots,p(+m)-p(-m)\big)\,,$$
where $m:=\left\lfloor\frac{n}{2}\right\rfloor$.  It is easy to see that $\varphi$ is linear.  It is also surjective since
$$\varphi(q_j)=e_j\,,$$
where $e_1,e_2,\ldots,e_m\in\mathbb{R}^m$ are the usual standard basis vectors, and
$$q_j(x):=\frac{\prod\limits_{i\in[m]\setminus\{j\}}\,\left(x^2-i^2\right)}{\prod\limits_{i\in[m]\setminus\{j\}}\,\left(j^2-i^2\right)}\in \mathbb{R}[x]\text{ for }j=1,2,\ldots,m\,.$$
Here, $[m]:=\{1,2,\ldots,m\}$.  
The next task is to show that $S=\ker(\varphi)$.  Clearly, by the definition of $S$, we see that $S\subseteq \ker(\varphi)$.  We need to prove that $\ker(\varphi)\subseteq S$ as well.  Let $p$ be an arbitrary element of $\ker(\varphi)$.  Define
$$f_p(x):=p(+x)-p(-x)\in\mathbb{R}[x]\,.$$
Thus, the roots of $f_p(x)$ are $0,\pm1,\pm2,\ldots,\pm m$.  Consequently, $f_p$ has at least $$2m+1=2\left\lfloor\frac{n}{2}\right\rfloor+1\geq 2\left(\frac{n-1}{2}\right)+1=n$$ roots.  As the degree of $f_p$ is at most the degree of $p$, which is $n-1$, we conclude that $f_p$ is identically zero.  That is, $p(+x)=p(-x)$ identically, whence $p(+\alpha)=p(-\alpha)$ for all $\alpha\in\mathbb{R}$, so that $p\in S$.

Alternatively, just observe that $S$ is spanned by the $\left\lceil\dfrac{m}{2}\right\rceil$ linearly independent polynomials $$1,x^2,x^4,\ldots,x^{2\,\left\lceil\frac{n}{2}\right\rceil-2}\,.$$  That is, every nonzero term in a polynomial $p\in S$ must have an even degree.
A: $\mathbf0$ is the zero polynomial, the null vector of the space $p_n(\mathbb{R})$, given with $\mathbf0(\alpha)=\sum_{i=0}^n0\cdot\alpha^i=0$ f.a. $\alpha$ as it works as the identity for pointwise addition of functions. Thus, especially $\mathbf0(\alpha)=0=\mathbf0(-\alpha)$ f.a. $\alpha\in\mathbb{R}$. Followingly, $\mathbf0\in S$.
Your work for the first two closure conditions look perfectly fine.
EDIT: A function $f$, i.e. in your case a polynomial of degree $n$(maximal), s.t. $f(x)=f(-x)$ is called an even function. Besides prominent examples like sine, especially all constant functions are even. (Can you see why?)
A: I claim that
$S = \{ p(x) \in P_n(x) \mid \forall \alpha \in \Bbb R, \; p(\alpha) = p(-\alpha) \} \tag 1$
is precisely the set
$E = \left \{ p(x) = \displaystyle \sum_0^{n - 1} p_i x_i \in P_n(x) \mid p_i = 0, i \; \text{odd} \right \}; \tag 2$
that is, 
$S = E, \tag 3$
which means the polynomials in $P_n(x)$ which satisfy $p(\alpha) = p(-\alpha)$ are precisely those in which occur only even powers of $x$; for the present purposes we count the zero polynomial $\mathbf 0$ as having only even-degree terms, which makes sense insofar as $\mathbf 0$ is the constant polynomial whose value is $0$ for every $\alpha$:
$\mathbf 0(\alpha) = 0, \forall \alpha \in \Bbb R; \tag 4$
furthermore, 
$\mathbf 0(\alpha) = 0 = \mathbf 0(-\alpha), \; \forall \alpha \in \Bbb R; \tag 5$
thus we see that
$\mathbf 0 \in S. \tag 6$
We may prove the assertion (3) as follows:  first off, it is easy to see that
$E \subset S, \tag 7$
since even degree terms are of the form $ax^{2i}$, $a \in \Bbb R$; we have
$a\alpha^{2i} = a(\alpha^2)^i = a((-\alpha)^2)^i = a(-\alpha)^{2i}; \tag 8$
since any element of $E$ is the sum of such expressions, we see that (7) must bind.
Now suppose
$p(x) \in S, \tag 9$
and write
$p(x) = \eta(x) + \omega(x), \tag{10}$
where $\eta(x)$ consists of the even-degree monomials comprising $p(x)$, and $\omega(x)$ the odd; then for any $\alpha \in \Bbb R$ we have
$\eta(\alpha) + \omega(\alpha) = p(\alpha) = p(-\alpha) = \eta(-\alpha) + \omega(-\alpha) = \eta(\alpha) - \omega(\alpha), \tag{11}$
whence
$2 \omega(\alpha) = 0 \Longrightarrow \omega(\alpha) = 0; \tag{12}$
since $\omega(\alpha) = 0$ for every $\alpha \in \Bbb R$, we conclude that
$\omega(x) = 0, \tag{13}$
whence
$p(x) = \eta(x) \tag{14}$
has only terms of even degree; thus we see that
$S \subset E, \tag{15}$
so in fact
$S = E \tag{16}$
as claimed.  Now $S = E$ is clearly a subspace of $P_n(x)$:  the sum of two polynomials with only even-degree monomials obviously satisfies the same criterion, as does any product of such a polynomial with a scalar; and $\mathbf 0 \in E$ as we have seen.  
And thats the $\alpha$ and $\omega$ of it!
