I'm looking at the following problem:
Let $X$ be a topological Hausdorff space and $$\mu:X\rightarrow mSpec \ C(X,\mathbb{R})$$ $$ \qquad \qquad \qquad x\mapsto \{f\in C(X,\mathbb{R})\ |\ f(x)=0\}.$$
Show that $Y\subset X$ is closed if and only if there is an ideal $a\subset C(X,\mathbb{R})$ such that $\mu(Y)=V(a)\cap mSpec \ C(X,\mathbb{R})$ with $$V(a):=\{p\in mSpec \ C(X,\mathbb{R})\ |\ a\subset p\}.$$
I believe to have proven one direction, but I'm not sure about it:
Let $Y \subset X$ and let $a\subset C(X,\mathbb{R})$ be an ideal such that $\mu(Y)=V(a)\cap \ mSpecC(X,\mathbb{R})$. Let $V(a) \cap mSpec \ C(X,\mathbb{R})=\{(m_i)_{i\in I}\}$ for some $m_i\in mSpec \ C(X,\mathbb{R})$. For every $m_i \in mSpec \ C(X,\mathbb{R})$ there is exactly one $x_i\in X$ ( I know this from the lecture). Therefore $\mu^{-1}(\{(m_i)_{i\in I}\})=\{(x_i)_{i\in I}\}=Y$, which is closed since a union of dots is closed in a hausdorff space. Is that correct?
For the other direction I got this: Let $Y\subset X$ be closed. Let $A\subset C(X,\mathbb{R})$. Since the sets $\{x\in X \ |\ f(x)\neq 0\}$ for $f\in C(X,\mathbb{R})$ give us a basis of the topology on $X$ (again, lecture), we can write $$Y=X\backslash \bigcup_{f\in A}\{x\in X\ | \ f(x)\neq 0\}=\bigcup_{f\in A}\{x\in X\ |\ f(x)=0 \}=:\{(x_i)_{i\in I}\}.$$ So $\mu(Y)=\{(m_i)_{i\in I}\}$ for some $m_i \in mSpec \ C(X,\mathbb{R})$. How can I now construct an ideal $a$ such that $V(a)=\{(m_i)_{i\in I}\}?$ Taking the intersection of the $(m_i)_{i\in I}$ is obviously not correct.
Can someone help me? Thanks in advance.