$C^*(X)\cap C^P_{\infty}(X)$ is an $e-$ideal of $C^*(X)$

Question

Definitions:

$X$ is a completely regular Hausdorff topological space ; $C(X)$ is the set of all continuous function from $X$ to $\mathbb R$ and $C^*(X)$ is the set of all real valued bounded continuous function on $X.$ For $f\in C^*(X)$

$$E_{\epsilon}(f)=\{ x\in X:|f(x)|\le \epsilon\}$$ and if $I$ is an ideal of $C^*(X)$, then $$E(I)=\{E_{\delta}(f) : f\in I,\delta \gt 0 \}$$ And the ideal $I$ is called $e-$ideal of $C^*(X)$ if $\ \ \ E_{\epsilon}(f)\in E(I)\forall \epsilon \gt 0$ implies $f\in I$. $$C^P_{\infty}(X)=\{ f\in C(X):\{x\in X : |f(x)|\ge {1\over n}\}\in P \quad \forall n\in \mathbb N\}$$ where $P$ is an ideal of closed sets of $X$ i.e. $$P=\{A\subset X|A \text{ is closed in X s.t. i) }A,B\in P \implies A\cup B\in P\text{ and ii) }A\in X,B\subset A \text{ with B is closed in }X \implies B\in P\}$$

Claim is that $C^*(X)\cap C^P_{\infty}(X)$ is an $e$-ideal of $C^*(X)$. I cannot see that. And moreover the definition of $e$-ideal is a confusing. Because if $E_{\epsilon}(f)\in E(I)$, does not that by definition of $E(I)$ mean that $f\in I$? Then every ideal is $e$-ideal.

I know I'm misunderstanding a lot. Please explain.

Answer 1

You asked about the relation between the notions of $e$-ideal and ideal.

And moreover the definition of $e$-ideal is a confusing. Because if $E_{\epsilon}(f)\in E(I)$, does not that by definition of $E(I)$ mean that $f\in I$? Then every ideal is $e$-ideal.

Not every ideal is an $e$-ideal. As a simple example let us take $X=[0,1]$ and the set of all functions which vanish on some neighborhood of $\frac12$. $$I=\{f\in C^*(X); \frac12 \in \operatorname{Int} Z(f)\} = \{f\in C^*(X); (\exists\varepsilon>0) x\in (\frac12-\varepsilon,\frac12+\varepsilon) \Rightarrow f(x)=0\}.$$

It is easy to see that if $f$ and $g$ are zero in the interval $(\frac12-\varepsilon,\frac12+\varepsilon)$ then so is $f-g$. And if $f$ is zero on this interval, then so is $fh$ for any $h$. So $I$ is indeed an ideal.

On the other hand, let us consider the function $f(x)=x-\frac12$. Clearly, $f\notin I$. But for any $\varepsilon>0$ we have $$E_\varepsilon(f)=(\frac12-\varepsilon,\frac12+\varepsilon).$$ And there is function $g\in I$ such that $E_\varepsilon(g)=E_\varepsilon(f)$. We can take a piecewise linear function which is zero on the interval $(\frac{1-\varepsilon}2,\frac{1+\varepsilon}2)$ and attains the value $\varepsilon$ exactly at the points $\frac12\pm\varepsilon$ (and outside this interval it has values at least $\varepsilon$). $$g(x)= \begin{cases} 0 & \text{if }|x-\frac12|<\frac\varepsilon2, \\ \varepsilon & \text{if }|x-\frac12|>\varepsilon, \\ 2x-\varepsilon & \text{otherwise}. \end{cases} $$ We see that $E_\varepsilon(f)\in E(I)$ for all $\varepsilon>0$.

Basically the same example is given in part 4 of Problem 2L (where the notion of an $e$-ideal is defined) in Gillman L., Jerison M. Rings of continuous functions (Princeton, 1960).

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Claim is that $C^*(X)\cap C^P_{\infty}(X)$ is an $e$-ideal of $C^*(X)$. I cannot see that.

I will copy here the relevant part of the proof of Theorem 2.1 from the paper S. K. Acharyya and S. K. Ghosh: Functions in $C(X)$ with support lying on a class of subsets of $X$, Topology Proceedings, Volume 35 (2010), pages 127-148, http://topology.auburn.edu/tp/reprints/v35/tp35010.pdf If there are still unclear parts, let me know in the comments.

Let $I = C^{\mathcal P}_\infty(X)\cap C^*(X)$ and $E_\varepsilon(f)\in E(I)$ for all $\varepsilon>0$. Suppose $n\in\mathbb N$. Then there is $g\in I$ and $\varepsilon>0$ such that $E_{\frac1{n+1}}(f)=E_\delta(g)$. Thus $\{x\in X; |f(x)|\le\frac1{n+1}\}=\{x\in X; |g(x)|\le\delta\}$ and so $\{x\in X; |f(x)|\ge\frac1n\}\subseteq\{x\in X; |g(x)|\ge\delta\}$. Choose $m\in\mathbb N$ such that $\delta\ge\frac1m$. Then $\{x\in X; |f(x)|\ge\frac1n\}\subseteq\{x\in X; |g(x)|\ge\frac1m\}$. Since $g\in C^{\mathcal P}_\infty(X)$, we can say that $\{x\in X; |f(x)|\ge\frac1n\}$ is a member of $\mathcal P$. Consequently, $f\in I$ and hence $I$ is an $e$-ideal of $C^*(X)$.

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One more point, which I already mentioned in comment (and which I can remove if you edit your question). The way you describe $P$ does not make much sense - the way it is written in the current revision of your question, you are defining a set $P$ using the set $P$. If you look, for example, at the paper linked above:

$\mathcal P$ is an ideal of closed sets in $X$ means that $\mathcal P$ is a family of closed subsets of $X$ which fulfills these two conditions:

• If $A,B\in\mathcal P$, then $A\cup B\in\mathcal P$.
• If $A\in\mathcal P$ and $B$ is a closed subset of $X$ such that $B\subseteq A$, then $B\in\mathcal P$.

This is precisely the ideal in order-theoretic sense if we consider the set of all closed subsets of $X$ ordered by inclusion. It is also very similar to ideals in set theory, with the difference that we only consider closed sets.