<p>I am now learning <em>Game Theory</em> by Fudenberg and Tirole, and I am thinking whether Glicksberg's existence theorem is a generalization of Nash Existence Theorem.</p>
<blockquote>
<p><strong>Theorem 1.1 in FT (Nash)</strong>: Every finite strategic-form game has a mixed strategy equilibrium.</p>
</blockquote>
<blockquote>
<p><strong>Theorem 1.3 in FT (Glicksberg)</strong>: For a strategic form game, if the strategy spaces <span class="math-container">$S_{i}$</span> are non-empty compact subsets of a metric space and payoff functions <span class="math-container">$u_{i}$</span> are continuous, then there exists a Nash equilibrium in mixed strategies.</p>
</blockquote>
<p>Because the payoff functions <span class="math-container">$u_{i}$</span> in finite strategic-form games are just polynomials by definition, they are continuous. I just want to know whether the strategy spaces in finite games are compact because generally we don't specify the topology in a finite game.</p>
<p>For example, for the following game, <span class="math-container">\begin{matrix}&amp;a&amp;b\\L&amp;3,3&amp;0,0\\R&amp;0,0&amp;-1,-1\end{matrix}</span>
<span class="math-container">$(a,L)$</span> is the unique Nash Equilibrium, but we don't know what topology we impose on the strategy spaces.</p>
<p>In conclusion, I think Glicksberg's theorem is not a generalization of Nash's theorem since we don't know the topology on strategy spaces in some finite games, which means that we don't know whether they are compact and Glicksberg's theorem is not applicable, is my argument correct?</p>
<p>Thanks in advance!</p>

CC BY-SA 4.0
2021-12-26T14:27:55.987Z
I don't think so. Note that a space $X$ is said to be compact, if every open covering of $X$ contains a finite sub-collection that covers $X$. Thus, if the strategic form game is finite, then it is compact, since it can be covered by a finite open sub-collection. Compactness can be interpreted as indicating finiteness.
Holger I. Meinhardt
CC BY-SA 4.0
2021-12-26T14:56:13.407Z
There is only one metrizable topology on a finite set, the discrete one. This makes every finite set a compact metrizable space and Glicksberg's existence theorem generalizes  Nash's existence theorem.
Michael Greinecker
<game-theory>
Is Glicksberg's existence theorem a generalization of Nash Existence Theorem?
John
# Is Glicksberg's existence theorem a generalization of Nash Existence Theorem?

(Asked by: username_0 on 2021-12-26)

I am now learning *Game Theory* by Fudenberg and Tirole, and I am thinking whether Glicksberg's existence theorem is a generalization of Nash Existence Theorem.

> **Theorem 1.1 in FT (Nash)**: Every finite strategic-form game has a mixed strategy equilibrium.

> **Theorem 1.3 in FT (Glicksberg)**: For a strategic form game, if the strategy spaces $S\_{i}$ are non-empty compact subsets of a metric space and payoff functions $u\_{i}$ are continuous, then there exists a Nash equilibrium in mixed strategies.

Because the payoff functions $u\_{i}$ in finite strategic-form games are just polynomials by definition, they are continuous. I just want to know whether the strategy spaces in finite games are compact because generally we don't specify the topology in a finite game.

For example, for the following game, \begin{matrix}&a&b\\L&3,3&0,0\\R&0,0&-1,-1\end{matrix}
$(a,L)$ is the unique Nash Equilibrium, but we don't know what topology we impose on the strategy spaces.

In conclusion, I think Glicksberg's theorem is not a generalization of Nash's theorem since we don't know the topology on strategy spaces in some finite games, which means that we don't know whether they are compact and Glicksberg's theorem is not applicable, is my argument correct?

Thanks in advance!

**Tags:** game-theory

> **Comments:**
> **username_1**: I don't think so. Note that a space $X$ is said to be compact, if every open covering of $X$ contains a finite sub-collection that covers $X$. Thus, if the strategic form game is finite, then it is compact, since it can be covered by a finite open sub-collection. Compactness can be interpreted as indicating finiteness.  
> **username_2**: There is only one metrizable topology on a finite set, the discrete one. This makes every finite set a compact metrizable space and Glicksberg's existence theorem generalizes  Nash's existence theorem.  

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## 0 answers

