bmtc-133-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

1. Which of the following statements are true or false? Give reasons for your answers in the form of a short proof or counter-example, whichever is appropriate:
   i) Every infinite set is an open set.

Background

1. Open Set: In topology, an open set is a set that, for every point in the set, includes some neighborhood of that point within the set. The exact definition of "neighborhood" depends on the topology defined on the space.
2. Topological Space: A topological space is a set equipped with a topology, a collection of open sets that satisfies certain properties (such as the union of open sets is open, and the finite intersection of open sets is open).

Analysis

Counterexample

• Take the set $S=\mathbb{Z}$$S=\mathbb{Z}$S=ZS = \mathbb{Z}$S=\mathbb{Z}$, the set of all integers. This is an infinite set.
• However, $S$$S$SS$S$ is not an open set in $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ with the standard topology. For any integer $z\in \mathbb{Z}$$z\in \mathbb{Z}$z inZz \in \mathbb{Z}$z\in \mathbb{Z}$, there is no open interval around $z$$z$zz$z$ that is entirely contained within $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$, as any open interval around $z$$z$zz$z$ will contain non-integer real numbers.

Conclusion

Background

1. Logical AND ($\wedge$$\wedge$^^\wedge$\wedge$): The statement $p\wedge q$$p\wedge q$p^^qp \wedge q$p\wedge q$ is true if and only if both $p$$p$pp$p$ and $q$$q$qq$q$ are true.
2. Negation ($\sim$$\sim$∼\sim$\sim$): The negation $\sim p$$\sim p$∼p\sim p$\sim p$ is true if and only if $p$$p$pp$p$ is false.
3. Logical Implication ($\to$$\to$rarr\rightarrow$\to$): The implication $p\to q$$p\to q$p rarr qp \rightarrow q$p\to q$ is true if either $p$$p$pp$p$ is false or $q$$q$qq$q$ is true (or both).

Analysis

1. Original Statement: $p\wedge \sim q$$p\wedge \sim q$p^^∼qp \wedge \sim q$p\wedge \sim q$ is true if and only if $p$$p$pp$p$ is true and $q$$q$qq$q$ is false.
2. Negation of the Original Statement: The negation of $p\wedge \sim q$$p\wedge \sim q$p^^∼qp \wedge \sim q$p\wedge \sim q$ is true if and only if either $p$$p$pp$p$ is false or $q$$q$qq$q$ is true. This is denoted as $\sim (p\wedge \sim q)$$\sim (p\wedge \sim q)$∼(p^^∼q)\sim (p \wedge \sim q)$\sim (p\wedge \sim q)$.
3. Comparing with $p\to q$$p\to q$p rarr qp \rightarrow q$p\to q$: The statement $p\to q$$p\to q$p rarr qp \rightarrow q$p\to q$ is true if either $p$$p$pp$p$ is false or $q$$q$qq$q$ is true. This is exactly the condition for the negation of $p\wedge \sim q$$p\wedge \sim q$p^^∼qp \wedge \sim q$p\wedge \sim q$.

Conclusion

Background

1. Limit Point: A point $x$$x$xx$x$ is a limit point of a set $S$$S$SS$S$ in a topological space if every neighborhood of $x$$x$xx$x$ contains at least one point of $S$$S$SS$S$ different from $x$$x$xx$x$ itself.
2. Interval $(1-2,1]$$(1-2,1]$(1-2,1](1 – 2, 1]$(1-2,1]$: The interval $(1-2,1]$$(1-2,1]$(1-2,1](1 – 2, 1]$(1-2,1]$ is equivalent to $(-1,1]$$(-1,1]$(-1,1](-1, 1]$(-1,1]$, which includes all real numbers greater than $-1$$-1$-1-1$-1$ and less than or equal to $1$$1$11$1$.

Analysis

1. Neighborhood of $-1$$-1$-1-1$-1$: A neighborhood of $-1$$-1$-1-1$-1$ is any interval that contains $-1$$-1$-1-1$-1$. For example, consider the neighborhood $(-1-ϵ,-1+ϵ)$$(-1-ϵ,-1+ϵ)$(-1-epsilon,-1+epsilon)(-1 – \epsilon, -1 + \epsilon)$(-1-ϵ,-1+ϵ)$ for any $ϵ>0$$ϵ>0$epsilon > 0\epsilon > 0$ϵ>0$.
2. Points in the Interval: The interval $(-1,1]$$(-1,1]$(-1,1](-1, 1]$(-1,1]$ contains points arbitrarily close to $-1$$-1$-1-1$-1$ but does not include $-1$$-1$-1-1$-1$ itself. For any $ϵ>0$$ϵ>0$epsilon > 0\epsilon > 0$ϵ>0$, there are points in $(-1,1]$$(-1,1]$(-1,1](-1, 1]$(-1,1]$ that lie within the neighborhood $(-1-ϵ,-1+ϵ)$$(-1-ϵ,-1+ϵ)$(-1-epsilon,-1+epsilon)(-1 – \epsilon, -1 + \epsilon)$(-1-ϵ,-1+ϵ)$ of $-1$$-1$-1-1$-1$.

Conclusion

Background

1. Integrable Function: A function is said to be integrable (in the Riemann sense) if the Riemann integral of the function over a certain interval exists. This means that it is possible to calculate the area under the curve of the function over that interval.
2. Continuous Function: A function is continuous if, intuitively, you can draw its graph without lifting your pen from the paper. Formally, a function $f$$f$ff$f$ is continuous at a point $c$$c$cc$c$ if the limit of $f(x)$$f(x)$f(x)f(x)$f(x)$ as $x$$x$xx$x$ approaches $c$$c$cc$c$ is equal to $f(c)$$f(c)$f(c)f(c)$f(c)$.

Analysis

Counterexample

Conclusion

Background

1. Differentiable Function: A function $f$$f$ff$f$ is differentiable at a point $x=a$$x=a$x=ax = a$x=a$ if the derivative $f^{\prime }(a)$$f^{\prime }(a)$f^(‘)(a)f'(a)$f^{\prime }(a)$ exists. This means that the limit of the difference quotient $\frac{f(x)-f(a)}{x-a}$$\frac{f(x)-f(a)}{x-a}$(f(x)-f(a))/(x-a)\frac{f(x) – f(a)}{x – a}$\frac{f(x)-f(a)}{x-a}$ as $x$$x$xx$x$ approaches $a$$a$aa$a$ exists and is finite.
2. Absolute Value Function: The absolute value function $|x|$$|x|$|x||x|$|x|$ is not differentiable at $x=0$$x=0$x=0x = 0$x=0$ because it has a sharp corner at that point. The derivative from the left does not equal the derivative from the right.

Analysis

1. For $x<2$$x<2$x < 2x < 2$x<2$: $f(x)=-(x-2)+(3-x)=-2x+5$$f(x)=-(x-2)+(3-x)=-2x+5$f(x)=-(x-2)+(3-x)=-2x+5f(x) = -(x – 2) + (3 – x) = -2x + 5$f(x)=-(x-2)+(3-x)=-2x+5$.
2. For $2\le x<3$$2\le x<3$2 <= x < 32 \leq x < 3$2\le x<3$: $f(x)=(x-2)+(3-x)=1$$f(x)=(x-2)+(3-x)=1$f(x)=(x-2)+(3-x)=1f(x) = (x – 2) + (3 – x) = 1$f(x)=(x-2)+(3-x)=1$.
3. For $x\ge 3$$x\ge 3$x >= 3x \geq 3$x\ge 3$: $f(x)=(x-2)-(3-x)=2x-5$$f(x)=(x-2)-(3-x)=2x-5$f(x)=(x-2)-(3-x)=2x-5f(x) = (x – 2) – (3 – x) = 2x – 5$f(x)=(x-2)-(3-x)=2x-5$.

Conclusion