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definition of quotient topology

Muon

New member
May 15, 2017
2
I'm wondering, shouldn't
$\tau_Y=\left\{U\subseteq Y:\bigcup U =\left(\bigcup_{ {[a]\in U} }[a]\right)\in\tau_X\right\}$
be written
$\tau_Y=\left\{U\subseteq Y:\bigcup U =\left(\bigcup_{ {[a]\in U} }\left\{[a]\right\}\right)\in\tau_X\right\}$
instead?
since $\bigcup U=\bigcup_{ [a]\in U }\left\{[a]\right\}$
or is this not correct?
Wolfram mathworld writes it as $\bigcup_{ [a]\in U } a$ which doesn't make any sense to me. I've never really felt comfortable with these formulas.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,632
Leiden
I'm wondering, shouldn't
$\tau_Y=\left\{U\subseteq Y:\bigcup U =\left(\bigcup_{ {[a]\in U} }[a]\right)\in\tau_X\right\}$
be written
$\tau_Y=\left\{U\subseteq Y:\bigcup U =\left(\bigcup_{ {[a]\in U} }\left\{[a]\right\}\right)\in\tau_X\right\}$
instead?
since $\bigcup U=\bigcup_{ [a]\in U }\left\{[a]\right\}$
or is this not correct?
Wolfram mathworld writes it as $\bigcup_{ [a]\in U } a$ which doesn't make any sense to me. I've never really felt comfortable with these formulas.
Hi Muon , welcome to MHB,

Let's pick an example.

Suppose we pick $X=\{1,2\}$ and $\tau_X = \{\varnothing, X\} = \{\varnothing, \{1,2\}\}$.
And let's pick an equivalence relation such that $[1]=\{1,2\}$.
Then we get $Y=\{[1]\}$ and $\tau_Y = \{ \varnothing, Y \} = \{\varnothing, \{[1]\}\} = \{\varnothing, \{\{1,2\}\}\}$.

Let $U=Y=\{[1]\}$, then $\bigcup_{ [a]\in U }\left\{[a]\right\} = \{[1]\} = \{\{1,2\}\}$, but that's not an element of $\tau_X$ is it?
However, with the definition we get $\bigcup_{ {[a]\in U} }[a] = [1] = \{1,2\}$, which is indeed an element of $\tau_X$.
Wolfram mathworld seems to have a typo, since we'd get $1$ (although that's not even properly defined), which is not in $\tau_X$ either.
I can only assume that when they write $a$, when it is already established that $a$ is an equivalence class, that the equivalence class is intended, since $a$ on its own doesn't have any meaning.