Skip to main content
MathOverflow

Return to Question

edited title
Link
rmoron
rmoron
  • 21
  • 3

Why does normality imply that a countable base $B$ contains at least one set $U$ whose closure is a proper subset of another set $V \in B$?

deleted 7 characters in body
Source Link
Michael Albanese
Michael Albanese
  • 19.3k
  • 9
  • 87
  • 160

I'm reading Aliprantis and Border's excellent text, Infinite Dimensional Analysis: A Hitchhiker's Guide (PDF available at link, assuming I've done this properly), and I've reached an impasse in the proof of the Urysohn Metrization theorem (3.40, p. 91).

The premises give that $X$ is Hausdorff, and additionally under equivalent statement 3 that $X$ is regular and second countable, so it has a countable base $B$. These conditions also imply by a preceding theorem (2.49, p. 46) that $X$ is normal. For the purposes of the remainder of the proof (which is to show that $X$ can be embedded in the Hilbert cube given regularity and second countability), the authors construct a set $C$ of pairs of nested elements of $B$ such that: [ C = {(U,V):\bar{U}\subset V \ & \ U,V \in B} ]$C = \{(U,V):\bar{U}\subset V \ \& \ U,V \in B\}$. Then they say, "The normality of $X$ implies that $C$ is nonempty." I've spent considerable time and effort trying to verify this myself, and haven't been able to figure out. It's definitely not a preceding result in this textbook (which is generally very self-contained). And although I've looked around here on Mathoverflow, my topological background might be too weak to see how this is an easy corollary of another theorem.

I'm reading Aliprantis and Border's excellent text, Infinite Dimensional Analysis: A Hitchhiker's Guide (PDF available at link, assuming I've done this properly), and I've reached an impasse in the proof of the Urysohn Metrization theorem (3.40, p. 91).

The premises give that $X$ is Hausdorff, and additionally under equivalent statement 3 that $X$ is regular and second countable, so it has a countable base $B$. These conditions also imply by a preceding theorem (2.49, p. 46) that $X$ is normal. For the purposes of the remainder of the proof (which is to show that $X$ can be embedded in the Hilbert cube given regularity and second countability), the authors construct a set $C$ of pairs of nested elements of $B$ such that: [ C = {(U,V):\bar{U}\subset V \ & \ U,V \in B} ] Then they say, "The normality of $X$ implies that $C$ is nonempty." I've spent considerable time and effort trying to verify this myself, and haven't been able to figure out. It's definitely not a preceding result in this textbook (which is generally very self-contained). And although I've looked around here on Mathoverflow, my topological background might be too weak to see how this is an easy corollary of another theorem.

I'm reading Aliprantis and Border's excellent text, Infinite Dimensional Analysis: A Hitchhiker's Guide (PDF available at link, assuming I've done this properly), and I've reached an impasse in the proof of the Urysohn Metrization theorem (3.40, p. 91).

The premises give that $X$ is Hausdorff, and additionally under equivalent statement 3 that $X$ is regular and second countable, so it has a countable base $B$. These conditions also imply by a preceding theorem (2.49, p. 46) that $X$ is normal. For the purposes of the remainder of the proof (which is to show that $X$ can be embedded in the Hilbert cube given regularity and second countability), the authors construct a set $C$ of pairs of nested elements of $B$ such that $C = \{(U,V):\bar{U}\subset V \ \& \ U,V \in B\}$. Then they say, "The normality of $X$ implies that $C$ is nonempty." I've spent considerable time and effort trying to verify this myself, and haven't been able to figure out. It's definitely not a preceding result in this textbook (which is generally very self-contained). And although I've looked around here on Mathoverflow, my topological background might be too weak to see how this is an easy corollary of another theorem.

Source Link
rmoron
rmoron
  • 21
  • 3

Why does normality imply that a countable base $B$ contains at least one set $U$ whose closure is a proper subset of another set $V \in B$?

I'm reading Aliprantis and Border's excellent text, Infinite Dimensional Analysis: A Hitchhiker's Guide (PDF available at link, assuming I've done this properly), and I've reached an impasse in the proof of the Urysohn Metrization theorem (3.40, p. 91).

The premises give that $X$ is Hausdorff, and additionally under equivalent statement 3 that $X$ is regular and second countable, so it has a countable base $B$. These conditions also imply by a preceding theorem (2.49, p. 46) that $X$ is normal. For the purposes of the remainder of the proof (which is to show that $X$ can be embedded in the Hilbert cube given regularity and second countability), the authors construct a set $C$ of pairs of nested elements of $B$ such that: [ C = {(U,V):\bar{U}\subset V \ & \ U,V \in B} ] Then they say, "The normality of $X$ implies that $C$ is nonempty." I've spent considerable time and effort trying to verify this myself, and haven't been able to figure out. It's definitely not a preceding result in this textbook (which is generally very self-contained). And although I've looked around here on Mathoverflow, my topological background might be too weak to see how this is an easy corollary of another theorem.