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Let $X=\mathbb{R^2}$ and let

$$A=\{(0,0)\}\cup\{(x,y):x>0\}\subseteq X\;.$$

Notice that there is no compact set around $\{(0,0)\}$ because it will be open set. It looks like work for example of two locally compact space whose union are not locally compact. Is it true?

Brian M. Scott
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Gob
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    To write ${}$ please type \{\} – user251257 Aug 14 '16 at 14:48
  • Looks right to me – user251257 Aug 14 '16 at 15:03
  • Just note that compact sets can be open in general (example, consider space ${*}$ with any topology) - openness is a proper of a subset inside a topological space, while compactness is a property of an abstract topological space Y (here I am mean "abstract" in the sense that Y doesn't need to be thought of with an inclusion into another in order to talk of $Y$ being compact). However, once you fix this you will be correct. – Elle Najt Aug 14 '16 at 15:27
  • I misunderstood you. Could you clarify your comment again? What by fixing this? – Gob Aug 14 '16 at 15:39
  • Thank prof Brian M.Scott for editing my question.What is your comments about it. Dose work good? – Gob Aug 14 '16 at 16:27
  • Why should anyone assume the union of two locally compact sets would be locally compact? So, yes, it looks like you have found an example where this wouldn't be true. – fleablood Aug 14 '16 at 16:28

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