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Let $A$ be $\mathbb R$ under the upper limit topology (having as basis $\{(a,b] : a < b\}$), then the subspace $\{(x, -x)\mid x \text{ is irrational}\}$ of $A\times A$ is closed, uncountable and discrete.

I think the subspace is closed since singleton is closed in upper limit space so closed in the product space, and since irrational is uncountable then the subspace is uncountable.

But why it is discrete? Irrational as a subspace of upper limit space is not discrete why in the product space it is discrete? Thanks!

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  • $\begingroup$ you need an extra - in your subspace, see my answer. $\{(x,-x): x \text{ is irrational }\}$ instead would be correct, as it stands it's false. $\endgroup$ – Henno Brandsma Dec 27 '19 at 7:29
  • $\begingroup$ Edited, thanks! $\endgroup$ – Cathy Feb 5 at 13:14
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In fact in $A \times A$ it's the antidiagonal that is closed and discrete, and so are all its subsets, the antidiagonal being

$$C=\{(x,-x): x \in A\}$$ while the diagonal $$\Delta=\{((x,x):x \in A\}$$

is just homeomorphic to $A$ (this holds in any space), and as $A$ is far from discrete and the irrationals in $A$ too, the statement you're asking us to prove is false.

But for the subspace $C$ it is true, as $$C \cap \left((x-1,x] \times (-x-1,-x]\right) = \{(x,-x)\}$$ showing that each point of $C$ is an isolated point in the subspace topology on $C$. That $C$ is closed in $A \times A$ is also easy to see and hence all subspaces of $C$ are closed and relatively discrete in $A \times A$, the basis for all proofs of non-normality of $A \times A$ (that I know of).

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  • $\begingroup$ What is meant by $(x-1,x]\times(-x-1,-x]$? The set $$\{(y,z) \in A\times A : x-1<y\leqslant x, -x-1<z\leqslant -x\}\ \mathrm ? $$ $\endgroup$ – Math1000 Dec 27 '19 at 7:37
  • $\begingroup$ @Math1000 The standard product open set $\{(y,z) \in A \times A: x-1< y \le x, -x-1 < z \le -x\}$ (boundaries count on the right!) $\endgroup$ – Henno Brandsma Dec 27 '19 at 7:40
  • $\begingroup$ Wait, if this is true for all subsets of $C$, then what was the significance of "$x$ is irrational"? $\endgroup$ – Math1000 Dec 27 '19 at 7:49
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    $\begingroup$ @Math1000 There are two standard ideas in proving $A \times A$ non-normal: try to separate $\{(x,-x): x \text{ irrational }\}$ from $\{((x,-x): x \text{ rational }\}$ (Both are closed in $A \times A$) and using a Baire category argument to get a contradiction, or using essentially Jones' lemma as a counting argument contrasting with the separability of $A \times A$. The OP seems to have a text going for the former proof, I suppose. $\endgroup$ – Henno Brandsma Dec 27 '19 at 7:52
  • $\begingroup$ So indeed it was a typo, having $(x,x)$ instead of $(x,-x)$? I suppose I should delete my answer then... $\endgroup$ – Math1000 Dec 27 '19 at 8:01
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The topology on a space $X$ is discrete iff its diagonal $\Delta =\{(x,x):x\in X\}$ is open in the product space $X\times X$. For when the diagonal is open, then the topology on $X$ has the property that every singleton $\{x\}$ is open. Since every $(x,x)\in\Delta$ is contained in a basis element $\{U\times V: U,V\text{ open in }X\}$, and the only sets of the form $U\times V$ where $U$,$V$ open in $X$ and $(x,x)\in U\times V$ are $U=V=\{x\}$, so $(x,x)$ is open in $\Delta$. It follows that $\{x\}$ is open in $X$, from which we conclude that $X$ is discrete. And if $X$ is discrete then any set in $X\times X$ is open, so in particular $\Delta$ is open.

In $A$ the subspace $\{x:x\text{ is irrational}\}$ is clearly not open, for the set $\{x:x\text{ is rational}\}$ is dense in $A$ and so any neighborhood $U$ of $x$ would contain a rational number. But in $A\times A$, the diagonal $\Delta = \{(x,x): x\text{ is irrational}\}$ is open, as the only sets of the form $U\times V$ such that $(x,x)\in U\times V$ and $U,V\subset\{x:\text{ is irrational}\}$ are $U=V=\{x\}$, so $\{x\}\times \{x\}=\{(x,x)\}$ is open in $\Delta$, and singletons being open immediately implies that $\{(x,x):x\text{ is irrational}\}$ is discrete as a subspace of $A\times A$.

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