Post History
#10: Post edited
by
Michael Hardy
·
2026-06-14T22:23:36Z (3 months ago)
- Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another must be empty.
Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the topolgy whose basic open sets are all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the topology whose basic open sets are all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection- $\Big( (0.3,0.4)\times \left\{~0.35~\right\} \Big) \cup \Big( \left\{ ~0.35~\right\}\times(0.3,0.4) \Big)$
- contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
- Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another must be empty.
- Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the topology whose basic open sets are all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the topology whose basic open sets are all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection
- $\Big( (0.3,0.4)\times \left\{~0.35~\right\} \Big) \cup \Big( \left\{ ~0.35~\right\}\times(0.3,0.4) \Big)$
- contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
#9: Post edited
by
Michael Hardy
·
2026-06-14T22:23:14Z (3 months ago)
- Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another must be empty.
Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the set of all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the set of all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection- $\Big( (0.3,0.4)\times \left\{~0.35~\right\} \Big) \cup \Big( \left\{ ~0.35~\right\}\times(0.3,0.4) \Big)$
- contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
- Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another must be empty.
- Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the topolgy whose basic open sets are all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the topology whose basic open sets are all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection
- $\Big( (0.3,0.4)\times \left\{~0.35~\right\} \Big) \cup \Big( \left\{ ~0.35~\right\}\times(0.3,0.4) \Big)$
- contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
#8: Post undeleted
by
Michael Hardy
·
2026-06-14T21:47:59Z (3 months ago)
#7: Post edited
by
Michael Hardy
·
2026-06-14T21:47:24Z (3 months ago)
#6: Post deleted
by
Michael Hardy
·
2026-06-13T22:01:11Z (3 months ago)
#5: Post edited
by
Michael Hardy
·
2026-06-13T21:52:46Z (3 months ago)
Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another is empty.- Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the set of all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the set of all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection
- $\Big( (0.3,0.4)\times \left\{~0.35~\right\} \Big) \cup \Big( \left\{ ~0.35~\right\}\times(0.3,0.4) \Big)$
- contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
- Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another must be empty.
- Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the set of all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the set of all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection
- $\Big( (0.3,0.4)\times \left\{~0.35~\right\} \Big) \cup \Big( \left\{ ~0.35~\right\}\times(0.3,0.4) \Big)$
- contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
#4: Post edited
by
Michael Hardy
·
2026-06-13T21:52:05Z (3 months ago)
- Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another is empty.
- Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the set of all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the set of all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection
$$\Big( (0.3,0.4)\times \left\{~0.35~ ight\} \Big) \cup \Big( \left\{ ~0.35~ ight\}\times(0.3,0.4) \Big)$$- contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
- Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another is empty.
- Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the set of all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the set of all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection
- $\Big( (0.3,0.4)\times \left\{~0.35~ ight\} \Big) \cup \Big( \left\{ ~0.35~ ight\}\times(0.3,0.4) \Big)$
- contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
#3: Post edited
by
Michael Hardy
·
2026-06-13T21:51:02Z (3 months ago)
#2: Post edited
by
Michael Hardy
·
2026-06-13T21:50:38Z (3 months ago)
- Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another is empty.
- Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the set of all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the set of all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection
- $$
\Big( (0.3,0.4)\times \{\,0.35\,\} \Big) \cup \Big( \{\,0.35\,\}\times(0.3,0.4) \Big)- $$
- contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
- Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another is empty.
- Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the set of all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the set of all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection
- $$
- \Big( (0.3,0.4)\times \left\{~0.35~\right\} \Big) \cup \Big( \left\{ ~0.35~\right\}\times(0.3,0.4) \Big)
- $$
- contains only the one point $(0.35,0.35),$ which is not an open set in either topology.
#1: Initial revision
by
Michael Hardy
·
2026-06-13T21:49:19Z (3 months ago)
Your last sentence makes me think you haven't understood your own question. There is no reason to think that the intersection of an open set in one topology and an open set in another is empty.
Consider the set $(0,1)\times(0,1).$ Let $\mathcal T_1$ be the set of all sets of the form $U\times\{\,x\,\}$ for some open $U\subseteq(0,1)$ and some $x\in(0,1),$ and let $\mathcal T_2$ be the set of all sets of the form $\{\,x\,\}\times U$ for some open $U\subseteq(0,1)$ and some $x\in(0,1).$ Both $\mathcal T_1$ and $\mathcal T_2$ are topologies. In $\mathcal T_1,$ every horizontal slice through the square is a connected component of the space, and in $\mathcal T_2$ the same is true of every vertical slice. But the intersection
$$
\Big( (0.3,0.4)\times \{\,0.35\,\} \Big) \cup \Big( \{\,0.35\,\}\times(0.3,0.4) \Big)
$$
contains only the one point $(0.35,0.35),$ which is not an open set in either topology.