Math Is Fun Forum

betterthangauss

Member

Registered: 2009-10-31

Posts: 11

Real analysis question involving closures and unions

Let

be a finite collection of sets in a metric space

and set

Then the closure of

, namely

, satisfies

The union of the closures of A_i's being a subset of the closure of B is easy to show. However, the reverse direction is tripping me up a bit. Here's how I argued it:

Assume

Then

is adherent to

Then for any neighborhood

of

, we have

i.e., for any neighborhood

of

, we have

Therefore, for any neighborhood

of

, there is an integer

,

, such that

Hence, there is an integer

,

, such that

.

Therefore,

Hence,

Is this argument fallacious? I ask, because I can't see where the finite part of the union comes into play. It seems like I could use this exact same argument to say the result holds in the countable case too, but there is an obvious counterexample to that:

Namely, setting

to the set

of the rational numbers and each

to the singleton set

where

(since

is countable).

Then the closure of each

is just the set

, since each

is just the singleton

.

, the set of real numbers, while

.

Hence,  the two sides are not equal.

Can someone show me where this first argument above was going bad? Thanks.

Last edited by betterthangauss (2009-11-10 22:00:46)

betterthangauss

Member

Registered: 2009-10-31

Posts: 11

Re: Real analysis question involving closures and unions

Let me just post images instead, so that it's easier to read:

QUESTION

MY ANSWER

NOTE: By J_n I mean the set

Last edited by betterthangauss (2009-11-10 22:25:41)

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Real analysis question involving closures and unions

The order of quantifiers is important.

Therefore, for any neighborhood

of

, there is an integer

,

, such that

This is true but the value of

many depend of the neighbourhood

.

What you want to show is the there exists an integer

such that for any neighbourhood of

of

betterthangauss

Member

Registered: 2009-10-31

Posts: 11

Re: Real analysis question involving closures and unions

The order of quantifiers is important.

Therefore, for any neighborhood

of

, there is an integer

,

, such that

This is true but the value of

many depend of the neighbourhood

.

What you want to show is the there exists an integer

such that for any neighbourhood of

of

Thanks. After taking a second look at the problem I saw a really easy way to prove it. Namely:

Therefore,

Note that

is the finite union of closed sets, so it is closed (this doesn't hold in the countable case).

Therefore, since

is the smallest closed set containing

, we have

Thanks again for showing me my error.