<p>I was going through the proof of existence of a Nash Equilibria in finite normal form games (Proof via Brouwer’s theorem) and got a question regarding the requirement of finiteness for the number of <strong>actions</strong> a player can play. In particular, the proof hints at the update equations where players "compute" their payoffs/utility of playing a best response, before going deeper into Brouwer's. I would think that, in this framework, the number of actions needs to be finite otherwise it would be impossible to compute those "utility-updates"... but is it the main point? Or am I missing something?</p>

<p>Edit: The proof I am looking at is: <a href="https://www.cs.ubc.ca/cgi-bin/tr/2007/TR-2007-25.pdf" rel="nofollow noreferrer">https://www.cs.ubc.ca/cgi-bin/tr/2007/TR-2007-25.pdf</a>. The update rules are in "Theorem 23".</p>

CC BY-SA 3.0
2018-04-21T22:57:46.057Z
Can you tell which proof you saw where? There is more than one way to prove the result via Brouwer's fixed point theorem.
Michael Greinecker
CC BY-SA 3.0
2018-04-22T08:37:45.257Z
I have added a link to a fac-simile proof.
gogu
CC BY-SA 3.0
2018-04-22T09:08:03.810Z
Well, there are two steps in the proof where finiteness of action spaces is used. First, the Brouwer fixed-point theorem is for finite dimensional spaces only. There are infinite-dimensional versions of the Brouwer fixed point theorem, but guaranteeing continuity and compactness is not straightforward in this case. If you would have a continuum of strategies, you get the problem that the probability of all actions might well be zero, so you cannot handle probabilities "pointwise".
Michael Greinecker
CC BY-SA 3.0
2018-04-22T09:08:06.900Z
But the existence of Nash equilibria works even without finiteness by work of Fan and Glicksberg. If action spaces are compact (metric spaces) and payoff function continuous, a Nash equilibrium will exist.
Michael Greinecker
<game-theory><nash-equilibrium>
Existence of nash equilibria in finite games
gogu
# Existence of nash equilibria in finite games

(Asked by: username_0 on 2018-04-21)

I was going through the proof of existence of a Nash Equilibria in finite normal form games (Proof via Brouwer’s theorem) and got a question regarding the requirement of finiteness for the number of **actions** a player can play. In particular, the proof hints at the update equations where players "compute" their payoffs/utility of playing a best response, before going deeper into Brouwer's. I would think that, in this framework, the number of actions needs to be finite otherwise it would be impossible to compute those "utility-updates"... but is it the main point? Or am I missing something?

Edit: The proof I am looking at is: <https://www.cs.ubc.ca/cgi-bin/tr/2007/TR-2007-25.pdf>. The update rules are in "Theorem 23".

**Tags:** game-theory, nash-equilibrium
> **Comments:**
> **username_1**: Can you tell which proof you saw where? There is more than one way to prove the result via Brouwer's fixed point theorem.  
> **username_0**: I have added a link to a fac-simile proof.  
> **username_1**: Well, there are two steps in the proof where finiteness of action spaces is used. First, the Brouwer fixed-point theorem is for finite dimensional spaces only. There are infinite-dimensional versions of the Brouwer fixed point theorem, but guaranteeing continuity and compactness is not straightforward in this case. If you would have a continuum of strategies, you get the problem that the probability of all actions might well be zero, so you cannot handle probabilities "pointwise".  
> **username_1**: But the existence of Nash equilibria works even without finiteness by work of Fan and Glicksberg. If action spaces are compact (metric spaces) and payoff function continuous, a Nash equilibrium will exist.  

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## 0 answers

