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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?

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    $\begingroup$ 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. $\endgroup$ – JustDroppedIn Apr 19 at 0:27
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The assumption of unique fixed points might be considered a red herring.

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.

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