Timeline for Approximation of Inductive Tensor Product $C(X) \bar{\otimes} C(Y)$
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 24, 2020 at 23:54 | comment | added | Sanae Kochiya | Thank you for your advice, @NikWeaver. Now I finally realize when finding basis I should start from leaving all but one coordinate $0$ but I can not imagine I missed this …. | |
| Jan 23, 2020 at 23:59 | comment | added | Nik Weaver | No ... no. I really think you should start by trying to understand the finite dimensional case. | |
| Jan 23, 2020 at 1:39 | comment | added | Sanae Kochiya | @NikWeaver I noticed where my mistake is .... In general your statement is right. When one of $X$ or $Y$ is infinite, $\vert C(X) \oplus C(Y) \vert$ is smaller than $\vert C(X \times Y) \vert$. I made a silly mistake here. $f_x \circ P_x$ is continuous in $X \times Y$ whenever $f_x \in C(X)$ but not the other way. | |
| Jan 22, 2020 at 17:45 | comment | added | Nik Weaver | You are wrong. Consider the case where $X$ and $Y$ are both singletons. | |
| Jan 22, 2020 at 17:19 | comment | added | Sanae Kochiya | @NikWeaver Now for set bijiection between $C(X \times Y)$ and $C(X) \oplus C(Y)$, you can refer to this link: math.stackexchange.com/questions/3516784. In $X \times Y$ (equipped with product topology), let $P_x, P_y$(resp.) be projection on $X, Y$(resp.). For any $f \in C(X \times Y)$, I consider the mapping $f \mapsto (f \circ P_x, f \circ P_y)$. The other direction is similar (please check the links for more details). | |
| Jan 22, 2020 at 16:32 | comment | added | Sanae Kochiya | @NikWeaver Assume both $X$ and $Y$ are finite sets. Let's fix $x_1 \in X$. For any two different $y_i, y_j \in Y$, ($\chi_{\{x_1\}}, \chi_{\{y_i\}}$) is linearly independent with ($\chi_{\{x_1\}}, \chi_{\{y_j\}}$).Then in a basis of $C(X) \oplus C(Y)$, at least we have $\vert Y \vert$ many linearly independent elements. For any two different $x_n, x_m \in X$ and a fixed $y \in Y$, also we have ($\chi_{\{x_n\}}, \chi_{\{y\}}$) is independent with ($\chi_{\{x_m\}}, \chi_{\{y\}}$). If there are no mistakes above, then the basis size of $C(X) \times C(Y)$ will be $\vert X \vert$*$\vert Y \vert$. | |
| Jan 22, 2020 at 4:08 | comment | added | erz | For nice space $X$ and $Y$, $C(X\times Y)=C(X,C(Y))$. | |
| Jan 21, 2020 at 14:15 | review | Close votes | |||
| Feb 5, 2020 at 3:05 | |||||
| Jan 21, 2020 at 11:52 | comment | added | Nik Weaver | The vector space dimension of $C(X) \oplus C(Y)$ is the sum of $|X|$ and $|Y|$, the dimension of $C(X\times Y)$ is their product. | |
| Jan 21, 2020 at 6:10 | history | edited | Sanae Kochiya | CC BY-SA 4.0 |
added 118 characters in body
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| Jan 21, 2020 at 6:03 | comment | added | Sanae Kochiya | @YemonChoi I am confused by the difference between the terminology used in my book and results from Google. These two terms mean the same but in case someone will check the book I will follow what the book is using. | |
| Jan 21, 2020 at 5:59 | comment | added | Sanae Kochiya | @NikWeaver I believe your concern is correct. Here I should change the $\times$ to $\oplus$ because I am taking direct sum of two Banach spaces (i,e, ($C(X), \|\cdot\|_{\infty}$ and ($C(Y), \|\cdot\|_{\infty}$). When $X$ and $Y$ are finite, dim$C(X)$ = $\vert X \vert$ because in this case $C(X)$ is the span of characteristic function of singleton. Hence dim[$C(X) \oplus C(Y)$] will be equal to dim$C(X \times Y)$. | |
| Jan 21, 2020 at 5:05 | comment | added | Nik Weaver | What is $C(X) \times C(Y)$? The cartesian product? If so, it can't be homeomorphic to $C(X \times Y)$. Look at the case where $X$ and $Y$ are finite, the dimensions don't match. | |
| Jan 21, 2020 at 4:20 | comment | added | Yemon Choi | Regarding terminology: I think this is usually called the injective tensor product of Banach spaces, not the "inductive" tensor product | |
| Jan 21, 2020 at 3:55 | review | First posts | |||
| Jan 21, 2020 at 5:32 | |||||
| Jan 21, 2020 at 3:53 | history | asked | Sanae Kochiya | CC BY-SA 4.0 |