# Why Compactness is Necessary at Minimax Theorem

According to Von Neumann's minimax theorem, I have

$$\max_{x\in X} \min_{y\in Y}f(x,y)=\min_{y\in Y} \max_{x\in X} f(x,y)$$

for some compact sets $X$ and $Y$ and a convex (in $y$), concave (in $x$) function $f(x,y)$. I wonder why I need the compactness to claim that this holds?

Whatever function I choose from $X$ and $Y$, $f(x,y)$ will be concave in $x$ and convex in $y$, which suggests a saddle point, that implies minimax theorem.

Thank you very much

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I am not sure how you conclude so easily that there will be a saddle point? I could not show you off-hand where precisely it is needed, but, from memory, vNM originally used Brouwer's Fixed Point theorem (which assume compactness); and later refinements always had compactness arguments. – gnometorule Jan 27 '13 at 0:26
I am not able to conclude eventually but it just seems so. Assuming that I have no sets but just a function $f(x,y)$, then, if the function is concave in $x$ and convex in $y$, then there exists a saddle point. Now I only assume that I have collection of $x$ and $y$ saved in some sets $X$ and $Y$ and which ever I choose will give me a sadlde point. – Seyhmus Güngören Jan 27 '13 at 0:30
Just one more remark. I also know that Sion's minimax theorem drops the necessity of two compact sets to a single one. – Seyhmus Güngören Jan 27 '13 at 0:32
How do you guarantee the existence of the inner minima/maxima (when the outer variable is fixed) without compactness? – Jyrki Lahtonen Jan 27 '13 at 16:00
@JyrkiLahtonen hmm. So you say that whenever I fix say $y\in Y$ then if $X$ is not compact, I cannot claim that $f(x,y)$ will have indeed a maxima. Because for $X$ there will be some infinite sequence with non converging sub-sequence indicating no maxima? then It is also correct for the other way around when I fix $x\in X$ and to find the minima for $y$. This means both $X$ and $Y$ should be compact? but Sions minimax require only one set to be compact.. – Seyhmus Güngören Jan 27 '13 at 16:21