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# Prove that any closed subset of a locally compact set is locally compact?

locally closed set

### 3 Answers

- Awms ALv 77 years agoFavorite Answer
Let A be a closed subset of the locally compact space X.

Let x be an element of A. Then x is an element of X, and therefore has a compact neighborhood C. Then the intersection C n A is closed in the space C, which is compact, so C n A is compact.

Since x is an element of C n A, this shows x has a compact neighborhood; and since x in A was arbitrary, this shows A is locally compact.

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- cosanoLv 43 years ago
i think of it rather is in the incorrect area. in actuality i think of it rather is the incorrect place to invite this way of question. you like a greater extreme high quality board for topology questions. in spite of everything, i will help get you began. In my own opinion, i've got discovered that is easiest to pass approximately proving all those issues in case you first initiate out by utilising itemizing the definitions of each theory linked with the question. this helps you to work out what this is you could desire to play with. as an occasion, you will desire to understand that an area is Hausdorff in case you could build 2 disjoint open instruments around 2 disjoint factors in that area. this would desire to tell you that using fact being Hausdorff is a situation, you ought to use that sources in the evidence. it may additionally behoove you to determine the circumstances for being domestically compact. i don't understand precisely what past awareness you're allowed to apply for the reason which you have not pronounced it, yet you have probable considered someplace quite a few proofs approximately equivalent family members to saying a collection is domestically compact and Hausdorff. One such equivalent fact is that if an area X is domestically compact and Hausdorff, then each element in X has a closed, compact community. in simple terms play around with those data and that would desire to get you began on a evidence.

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