Skip to main content
MathOverflow

Return to Answer

replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Source Link
Community Bot
Community Bot
  • 1
  • 2
  • 3

I recently found an answer to a similar question. Suppose: $$(\forall V\in \mathcal U)(\exists W\in \mathcal U)(U\circ W\subseteq V\circ U)$$ By axiom of choice,for each $V \in \mathcal U$, there's some symmetric $D_V\in \mathcal U$ such that $$D_V\circ D_V\subseteq V$$

and there's some symmetric $W_V\in \mathcal U$ such that:

$$U\circ W_V \subseteq D_V\circ U$$ and $$W_V\subseteq D_V$$

so $$W_V\circ U\circ W_V \subseteq W_V\circ D_V\circ U\subseteq D_V\circ D_V\circ U\subseteq V\circ U$$

Therefore

$$\overline U=\bigcap_{W\in \mathcal U}W\circ U \circ W\subseteq \bigcap_{V\in \mathcal U}W_V\circ U \circ W_V\subseteq \bigcap_{V\in \mathcal U}V\circ U\subseteq U\circ U$$

Now see this threadthis thread.

I recently found an answer to a similar question. Suppose: $$(\forall V\in \mathcal U)(\exists W\in \mathcal U)(U\circ W\subseteq V\circ U)$$ By axiom of choice,for each $V \in \mathcal U$, there's some symmetric $D_V\in \mathcal U$ such that $$D_V\circ D_V\subseteq V$$

and there's some symmetric $W_V\in \mathcal U$ such that:

$$U\circ W_V \subseteq D_V\circ U$$ and $$W_V\subseteq D_V$$

so $$W_V\circ U\circ W_V \subseteq W_V\circ D_V\circ U\subseteq D_V\circ D_V\circ U\subseteq V\circ U$$

Therefore

$$\overline U=\bigcap_{W\in \mathcal U}W\circ U \circ W\subseteq \bigcap_{V\in \mathcal U}W_V\circ U \circ W_V\subseteq \bigcap_{V\in \mathcal U}V\circ U\subseteq U\circ U$$

Now see this thread.

I recently found an answer to a similar question. Suppose: $$(\forall V\in \mathcal U)(\exists W\in \mathcal U)(U\circ W\subseteq V\circ U)$$ By axiom of choice,for each $V \in \mathcal U$, there's some symmetric $D_V\in \mathcal U$ such that $$D_V\circ D_V\subseteq V$$

and there's some symmetric $W_V\in \mathcal U$ such that:

$$U\circ W_V \subseteq D_V\circ U$$ and $$W_V\subseteq D_V$$

so $$W_V\circ U\circ W_V \subseteq W_V\circ D_V\circ U\subseteq D_V\circ D_V\circ U\subseteq V\circ U$$

Therefore

$$\overline U=\bigcap_{W\in \mathcal U}W\circ U \circ W\subseteq \bigcap_{V\in \mathcal U}W_V\circ U \circ W_V\subseteq \bigcap_{V\in \mathcal U}V\circ U\subseteq U\circ U$$

Now see this thread.

added 17 characters in body
Source Link
user31967
user31967

I recently found an answer to a similar question. Suppose: $$(\forall V\in \mathcal U)(\exists W\in \mathcal U)(U\circ W\subseteq V\circ U)$$ By axiom of choice,for each $V \in \mathcal U$, there's some symmetric $D_V\in \mathcal U$ such that $$D_V\circ D_V\subseteq V$$

and there's some symmetric $W_V\in \mathcal U$ such that:

$$U\circ W_V \subseteq D_V\circ U$$ and $$W_V\subseteq D_V$$

so $$W_V\circ U\circ W_V \subseteq W_V\circ D_V\circ U\subseteq D_V\circ D_V\circ U\subseteq V\circ U$$

Therefore

$$\overline U=\bigcap_{W\in \mathcal U}W\circ U \circ W\subseteq \bigcap_{V\in \mathcal U}W_V\circ U \circ W_V\subseteq \bigcap_{V\in \mathcal U}V\circ U\subseteq U\circ U$$

Now see this thread.

I recently found an answer to a similar question. Suppose: $$(\forall V\in \mathcal U)(\exists W\in \mathcal U)(U\circ W\subseteq V\circ U)$$ By choice,for each $V \in \mathcal U$, some symmetric $D_V\in \mathcal U$ such that $$D_V\circ D_V\subseteq V$$

and there's some symmetric $W_V\in \mathcal U$ such that:

$$U\circ W_V \subseteq D_V\circ U$$ and $$W_V\subseteq D_V$$

so $$W_V\circ U\circ W_V \subseteq W_V\circ D_V\circ U\subseteq D_V\circ D_V\circ U\subseteq V\circ U$$

Therefore

$$\overline U=\bigcap_{W\in \mathcal U}W\circ U \circ W\subseteq \bigcap_{V\in \mathcal U}W_V\circ U \circ W_V\subseteq \bigcap_{V\in \mathcal U}V\circ U\subseteq U\circ U$$

Now see this thread.

I recently found an answer to a similar question. Suppose: $$(\forall V\in \mathcal U)(\exists W\in \mathcal U)(U\circ W\subseteq V\circ U)$$ By axiom of choice,for each $V \in \mathcal U$, there's some symmetric $D_V\in \mathcal U$ such that $$D_V\circ D_V\subseteq V$$

and there's some symmetric $W_V\in \mathcal U$ such that:

$$U\circ W_V \subseteq D_V\circ U$$ and $$W_V\subseteq D_V$$

so $$W_V\circ U\circ W_V \subseteq W_V\circ D_V\circ U\subseteq D_V\circ D_V\circ U\subseteq V\circ U$$

Therefore

$$\overline U=\bigcap_{W\in \mathcal U}W\circ U \circ W\subseteq \bigcap_{V\in \mathcal U}W_V\circ U \circ W_V\subseteq \bigcap_{V\in \mathcal U}V\circ U\subseteq U\circ U$$

Now see this thread.

Source Link
user31967
user31967

I recently found an answer to a similar question. Suppose: $$(\forall V\in \mathcal U)(\exists W\in \mathcal U)(U\circ W\subseteq V\circ U)$$ By choice,for each $V \in \mathcal U$, some symmetric $D_V\in \mathcal U$ such that $$D_V\circ D_V\subseteq V$$

and there's some symmetric $W_V\in \mathcal U$ such that:

$$U\circ W_V \subseteq D_V\circ U$$ and $$W_V\subseteq D_V$$

so $$W_V\circ U\circ W_V \subseteq W_V\circ D_V\circ U\subseteq D_V\circ D_V\circ U\subseteq V\circ U$$

Therefore

$$\overline U=\bigcap_{W\in \mathcal U}W\circ U \circ W\subseteq \bigcap_{V\in \mathcal U}W_V\circ U \circ W_V\subseteq \bigcap_{V\in \mathcal U}V\circ U\subseteq U\circ U$$

Now see this thread.