{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Rudin_2

# Rudin_2 - Basic Topology Written by Men-Gen Tsai email...

This preview shows pages 1–4. Sign up to view the full content.

Basic Topology Written by Men-Gen Tsai email: [email protected] 1. Prove that the empty set is a subset of every set. Proof: For any element x of the empty set, x is also an element of every set since x does not exist. Hence, the empty set is a subset of every set. 2. A complex number z is said to be algebraic if there are integers a 0 , ..., a n , not all zero, such that a 0 z n + a 1 z n - 1 + ... + a n - 1 z + a n = 0 . Prove that the set of all algebraic numbers is countable. Hint: For every positive integer N there are only finitely many equations with n + | a 0 | + | a 1 | + ... + | a n | = N. Proof: For every positive integer N there are only finitely many equa- tions with n + | a 0 | + | a 1 | + ... + | a n | = N. (since 1 n N and 0 ≤ | a 0 | ≤ N ). We collect those equations as C N . Hence C N is countable. For each algebraic number, we can form an equation and this equation lies in C M for some M and thus the set of all algebraic numbers is countable. 3. Prove that there exist real numbers which are not algebraic. Proof: If not, R 1 = { all algebraic numbers } is countable, a contra- diction. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4. Is the set of all irrational real numbers countable? Solution: If R - Q is countable, then R 1 = ( R - Q ) Q is countable, a contradiction. Thus R - Q is uncountable. 5. Construct a bounded set of real numbers with exactly three limit points. Solution: Put A = { 1 /n : n N } { 1 + 1 /n : n N } { 2 + 1 /n : n N } . A is bounded by 3, and A contains three limit points - 0, 1, 2. 6. Let E be the set of all limit points of a set E . Prove that S is closed. Prove that E and E have the same limit points. (Recall that E = E E .) Do E and E always have the same limit points? Proof: For any point p of X - E , that is, p is not a limit point E , there exists a neighborhood of p such that q is not in E with q = p for every q in that neighborhood. Hence, p is an interior point of X - E , that is, X - E is open, that is, E is closed. Next, if p is a limit point of E , then p is also a limit point of E since E = E E . If p is a limit point of E , then every neighborhood N r ( p ) of p contains a point q = p such that q E . If q E , we completed the proof. So we suppose that q E - E = E - E . Then q is a limit point of E . Hence, N r ( q ) where r = 1 2 min( r - d ( p, q ) , d ( p, q )) is a neighborhood of q and contains a point x = q such that x E . Note that N r ( q ) contains in N r ( p ) -{ p } . That is, x = p and x is in N r ( p ). Hence, q also a limit point of E . Hence, E and E have the same limit points. 2
Last, the answer of the final sub-problem is no. Put E = { 1 /n : n N } , and E = { 0 } and ( E ) = φ . 7. Let A 1 , A 2 , A 3 , ... be subsets of a metric space. (a) If B n = n i =1 A i , prove that B n = n i =1 A i , for n = 1 , 2 , 3 , ... (b) If B = i =1 , prove that B i =1 A i . Show, by an example, that this inclusion can be proper. Proof of (a): (Method 1) B n is the smallest closed subset of X that contains B n . Note that A i is a closed subset of X that contains B n , thus B n n i =1 A i .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 24

Rudin_2 - Basic Topology Written by Men-Gen Tsai email...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online