Is it a metric?

Q11. Basic Topology (Rudin)

For $x \in R^1$ and $y \in R^1$ define

(a) $d_1 (x, y) = (x - y)^2$

Determine whether it is a metric or not.

Solution: $d_1 (x, y) = (x - y)^2$ satisfies the condition that distance function is positive for all x, y and 0 when x = y since square is always non negative.

Also $d_1 (x, y) = d_1 (y, x)$ as $(x - y)^2 = (y-x)^2$

Finally we check the triangular inequality.  We have to show that $(x-y)^2 + (y-z)^2 \ge (x-z)^2$ . If this is true then we will have $2y^2 - 2xy - 2yz  + 2zx \ge  0 \implies (y-x)(y-z) \ge 0$ . But this is not true for all real numbers. For example if we have x < y < z then (y - x)(y - z) becomes negative. Thus x = 2, y = 4 and z = 7 violates the triangular inequality as $(2-4)^2 + (4-7)^2 \le (2-7)^2$.

Hence $d_1$ is not a metric.

(b) $d_2 (x, y) = \sqrt {|x -y| }$

Determine whether it is a metric or not.

Solution:  $d_2 (x, y) = \sqrt {|x -y| }$ satisfies the condition that distance function is positive for all x, y and 0 when x = y since square root of |x-y| is always non negative and equals 0 only when x=y.

Also $d_2 (x, y) = d_2 (y, x)$ as $\sqrt {|x -y|} = \sqrt {|y-x|}$

Finally we check the triangular inequality. We have to show that $\sqrt {|x -y|}  + \sqrt {|y-z|} \ge  \sqrt {|x-z|}$
 

Interior of a set

Q9. Basic Topology (Rudin)

Let $E^0$ denote the set of all interior points of a set E. 

(a) Prove that  $E^0$ is always open.

Solution: Suppose x is an arbitrary point in $E^0$. We will show that x is an interior point of $E^0$.

By definition of $E^0$,  x is an interior point of E. Hence there exists a neighborhood N of radius r of x, such that $N \subset E$.

Since N is a neighborhood therefore N is open. Hence all points of N are interior points of N i.e. they have neighborhoods that are subsets of N. Since N is a subset of E, hence those neighborhoods (of the points of N which are subsets of N) are also subsets of E. Hence all the points in N are interior points of E. Thus all the points of N are member of $E^0$ by definition (since $E^0$ contains all the interior point of E). Thus N is a subset of $E^0$. Since x is an arbitrary element of $E^0$, and it has a neighborhood which is a subset of $E^0$, hence x is interior point of $E^0$ and $E^0$ is open.

(b) Prove that E is open if and only if $E^0 = E$

Solution: If E is open, all point of E are interior points of E (by definition). Hence all points of E are in $E^0$ and no other (as $E^0$ contains only the interior points of E and all points of E are interior points). Hence $E^0 = E$

Again if $E^0 = E$ , E contains only the points present in $E^0$ and $E^0$ contains only the interior points of E. Thus all points in E are interior points. Hence E is open.

(c) If $G \subset E$ and G is open, prove that $G \subset E^0$

Solution: Since G is open, all points of G are interior points of G. Hence each point of G has neighborhood N which is a subset of G. As G is a subset of E, hence these N's are also subsets of E. Therefore for each point in G, there exist neighborhood N which is a subset of E. Hence all points in G are interior points of E. Since $E^0$ contains all interior points of E, $E^0$ contains all points of G. Hence $G \subset E^0$.

(d) Prove that the complement of $E^0$ is the closure of the complement of E.

Solution: Complement of $E^0$ has the points which are not interior points of E. Either they are not members of E which makes them member of $E^c$ or even if they are members of E, they are points whose   every neighborhood contains some elements which are not member of E. In other words complement of \whose every neighborhood contains some elements from $E^c$. Hence they are limit points of $E^c$.
Thus the complement of $E^0$ contains elements of $E^c$ and limit points of $E^c$ which makes it by definition closure of the complement of E.

(e) Do E and $\bar{E}$ always have same interiors?

Solution: No. Counterexample: E= Set of Rational Numbers. Interior of this set is Null Set for if q be any rational number, any neighborhood of q will contain irrationals hence it cannot be totally inside E. Limit Points of this set E consists of rationals as well as irrationals. Hence closure of the set of rational numbers is $\bar{E}$ (set of real numbers). $\bar{E}$ has plenty of interior points (in fact all points are interior points). Hence E and $\bar{E}$ does not have same interior.

(f) Do E and $E^0$ always have the same closures?

Solution:

Algebraic numbers are countable

Q2. Basic Topology (Rudin)

A complex number z is said to be algebraic if there are integers $a_0 , a_1 , \ldots , a_n$ , not all zero, such that

 $a_0 z^n + a_1 z^{n-1} + \ldots + a_{n-1} z + a_n = 0$ .

 Prove that the set of all algebraic numbers are countable. Hint: For every positive integer N there are only finitely many equations with $n +  | a_0 | + | a_1 | + \ldots +  | a_n | = N$

Solution: 

Set A is countable if we can establish one-one correspondence between A and set of positive integers \(\mathbb{N}\)

By Fundamental theorem of Algebra  $a_0 z^n + a_1 z^{n-1} + \ldots + a_{n-1} z + a_n = 0$  has exactly n complex roots.

Empty Set is a Subset of Every Set

Definition of 'subset' is: A is a subset of B if $x \in A \implies x \in B$
This is an implication statement (If P then Q where P and Q are propositions).
The truth table is Implication Statement is:
P   Q   P=>Q
T   T       T
T   F       F
F   T       T
F   F       T
where T is 'True' and F is 'False'.
That is the truth value of an implication statement is always 'True' when the first proposition (P) is false.
Consider the statement $x \in A \implies x \in B$
Suppose A is an empty set. Then 'x is in A' is false as A contains no element. Hence the implication statement's truth value is True.
Thus A is a subset of B. Since B is any set and A is an empty set, therefore empty set is a subset of any set.