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Show that the diagonal $\{(x,x): x\in X\}$ is closed in the metric space $(X\times X,d=\max\{d_X,d_X\})?$

My attempt:

Choose $(x,y)\in X\times X-\{(a,a): a\in X\}$

Then $c=d(x,y)/2>0.$ To show $B((x,y),c)\in X\times X-\{(a,a):a\in X\}$

If possible let $(h,h)\in B((x,y),c)$

Then $d(x,y)\le d(x,h)+d(y,h)<2c=d(x,y)!$

Am I correct?

dfeuer
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Sriti Mallick
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  • Yes, this is correct. – youler Jul 26 '13 at 03:31
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    Note that a closed diagonal is equivalent to your topological space having the Hausdorff property. http://www.proofwiki.org/wiki/Hausdorff_Space_iff_Diagonal_Set_on_Product_is_Closed – dls Jul 26 '13 at 03:33
  • In English, the first person, singular, subject pronoun is always capitalized. So no, "i" is not correct. – dfeuer Jul 26 '13 at 03:41
  • Out of curiosity: Isn't that true for any metric space? Or true in general for metrics that generate the product topology on $X\times X$? – Xena Jul 26 '13 at 04:53
  • @dfeuer: Not if you’re quoting some of e.e. cummings’s poetry. – Brian M. Scott Jul 26 '13 at 10:52

2 Answers2

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Looks good! IMO, the last line is kind of confusing in its brevity. You might want to label the steps, like so:

If possible let $(h,h)\in B((x,y),c)$. Then

\begin{align*}d(x,y)&\le d(x,h)+d(y,h)\quad\text{[triangle inequality]}\\ &\leq 2\cdot \max\{d(x,h),d(y,h)\}\quad\text{[definition of max]}\\ &= 2\cdot d((x,y),(h,h))\quad\text{[definition of product metric]}\\ &< 2\cdot c\quad\text{[assumption on $h$]}\\ &= d(x,y)\quad\text{[definition of $c$]} \end{align*}

Collapsing the inequalities, we get $d(x,y) < d(x,y)$, contradiction.

Souza
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Chris Culter
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The distance function $d: X \times X \longrightarrow \mathbb{R}$ is continuous. Hence, the diagonal $$\Delta := \{(x,x) ; x \in X\} = d^{-1}(\{0\})$$ is a closed set.

Souza
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Moacir Aloísio
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