# Closed graph and fixed points

I’m currently trying to understand the following Proposition from a paper i’m reading:

Prop.: Let $$X$$ and $$Y$$ be two Hausdorff topological linear spaces. Let $$H:X \times Y \rightarrow Y$$ be a continuous function with unique fixed point $$h(x) \in Y$$ for every $$x \in X$$, i.e. $$H(x,h(x))=h(x)$$ and for each $$x \in X$$ the fixed point $$h(x) \in Y$$ is unique. Then, graph $$h$$ is closed in the product topology.

Proof: Let $$(x_j,y_j)_{j \in J}\rightarrow (x,y)$$ be a converging net such that $$h(x_j)=y_j$$ for every $$j \in J$$. By continuity of $$H$$, it follows that $$H(x,y)-y=\lim_j[H(x_j,y_j)-y_j]=0$$, i.e. $$h(x)=y$$.

My question is: why do $$X$$ and $$Y$$ have to be topological vector spaces? Wouldn’t it be enough for them to be Hausdorff spaces?

• Well, for instance, you are considering a difference of the form $H(x,y)-y$. That automatically employs the vector space structure of $Y$, and moreover, the operations of $Y$ have to be continuous (hence $Y$ must be a topological vector space). I think $X$ can be an arbitary Hausdorff space though. – JustDroppedIn Apr 19 at 0:27

If we merely assume that $$X$$ is a topological space, $$Y$$ is a Hausdorff space, $$H: X\times Y\to Y$$ is continuous and we define $$G = \{ (x, y) \in X\times Y \mid H(x, y) = y \}$$ then we can already conclude that $$G$$ is closed in $$X\times Y$$, simply because it is the set on which $$H$$ agrees with the (also continuous) projection $$(x,y) \mapsto y$$.
If $$G$$ happens to be the graph of a function $$h: X\to Y$$, that may be useful for other reasons, but it does not change the fact that $$G$$ is closed.