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Let $\mathbb{R}^2 := \mathbb{R} \times \mathbb{R}$ have the product topology (where $\mathbb{R}$ has the standard order topology). Let $D = \{(x,x) | x \in \mathbb{R}\}$ be the diagonal in $\mathbb{R}^2$. Show that $D$ is closed in $\mathbb{R}^2$.

Here, its all in Reals. I did it using Hausdorff but I cant use that. My prof said show that using if its complement (i.e. $D^C$) is open then $D$ is closed or using some other technique.

Below is what I did, please check and let me know if there is anything wrong or if it require a better notation. Thanks!!

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Math_Is_Fun
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2 Answers2

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It's true for Hausdorff spaces but got the reals you can give a quite easy argument if you know that open segments like $(a,\rightarrow) =\{x \in \Bbb R: x > a\}$ and $(\leftarrow,a)=\{x \in \Bbb R: x < a\}$ are open:

Suppose $(x,y) \notin D$ so $x \neq y$. Suppose WLOG that $x < y$ and pick $z$ so that $x < z < y$, then $(\leftarrow,z) \times (z,\rightarrow)$ is open in $\Bbb R^2$ (in the product topology), contains $(x,y)$ and misses $D$: otherwise $(w,w) \in (\leftarrow,z) \times (z,\rightarrow)$ would imply $w < z$ and $w > z$ which cannot both be true.

Henno Brandsma
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  • I got your proof, I have few questions to ask you abt open segments, whats the difference b/w open intervals and segments? We didn't have this in our notes we only did open intervals, half open intervals and closed intervals. Does a segment have similar meaning as interval ? – Math_Is_Fun Mar 01 '20 at 04:08
  • @Math_Is_Fun an open segment has only one bound: it’s of the form ${x\mid x < a}$ or ${x\mid x > a}$ for some $a$ in the ordered set. An interval has two bounds, say $a< b$, and consists of all points in between. – Henno Brandsma Mar 01 '20 at 08:05
  • ok, got it now. :) – Math_Is_Fun Mar 01 '20 at 08:07
  • @Math_Is_Fun The definition of the order topology is that all open segments are open, open intervals are then also open as $(a,b)=(a, \rightarrow) \cap (\leftarrow,b)$ etc. a min or max of an ordered set cannot lie in an open interval, but does lie in open segments of one of the types. Hence their slightly different local bases. – Henno Brandsma Mar 01 '20 at 08:21
  • yeah, order topology says that its generated from a sub-basis that consists of union of positive and negative open rays, and wat u did here is an open interval. Anyway, would u mind checking my proof? I updated this Q with my proof, you can see that on the top. Kindly let me know if it needs a better notation or if there is something wrong. – Math_Is_Fun Mar 01 '20 at 08:31
  • @Math_Is_Fun Your proof is fine. You can suffice with the open segments as I did. No need for $\varepsilon$ and disjoint intervals. I only need that there is a $z$ between $x$ and $y$. It generalises to all ordered spaces (if no such $z$ exists, we take slightly different open segments (or open rays, as they're also called). I like concise and to the point proofs. – Henno Brandsma Mar 01 '20 at 08:39
  • ok, what if I use $\epsilon$, will it be a matter of grave concern? – Math_Is_Fun Mar 01 '20 at 08:42
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You sort of need to use that $\mathbb{R}^2$ is Hausdorff because $X$ is Hausdorff iff the diagonal in $X^2$ is closed.

I guess you are to show essentially this equivalence. You can do this by showing the compliment is open.

See this SE post

user722227
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