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This is the regular calculus sequence in the department. Students entering this sequence are to have mastered appropriate precalculus material and, in many cases, have had some previous experience with calculus in high school or elsewhere. All Autumn Quarter offerings of MATH 15100, 15200, and 15300 begin with a rigorous treatment of limits and limit proofs. Students may not take the first two quarters of this sequence for P/F grading. MATH 15100-15200 meets the general education requirement in mathematical sciences.
Let $A=\mathbb{Q} \cap(0, \infty)$. Find $A^{\circ}, A^{\prime}$, and $\bar{A}$.
(a) First we show that $\overline{A^c} \subseteq A^{\circ c}$. If $x \in \overline{A^c}$, then either $x \in A^c$ or $x$ is an accumulation point of $A^c$. If $x \in A^c$, then $x$ is certainly not in the interior of $A$; that is, $x \in A^{\circ c}$. On the other hand, if $x$ is an accumulation point of $A^c$, then every $\epsilon$-neighborhood of $x$ contains points of $A^c$. This means that no $\epsilon$-neighborhood of $x$ lies entirely in $A$. So, in this case too, $x \in A^{\circ c}$.
For the reverse inclusion suppose $x \in A^{\circ c}$. Since $x$ is not in the interior of $A$, no $\epsilon$-neighborhood of $x$ lies entirely in $A$. Thus either $x$ itself fails to be in $A$, in which case $x$ belongs to $A^c$ and therefore to $\overline{A^c}$, or else every $\epsilon$-neighborhood of $x$ contains a point of $A^c$ different from $x$. In this latter case also, $x$ belongs to the closure of $A^c$.
The function $f: \mathbb{R} \rightarrow \mathbb{R}: x \mapsto 5 x-8$ is continuous.
Let $a \in \mathbb{R}$. Given $\epsilon>0$, choose $\delta=\epsilon / 5$. If $|x-a|<\delta$, then $|f(x)-f(a)|=5|x-a|<5 \delta=\epsilon$.
$A$ set $A \subseteq \mathbb{R}$ is connected if and only if the only subsets of $A$ which are both open in $A$ and closed in $A$ are the null set and $A$ itself.
Suppose there exists a nonempty set $U$ which is properly contained in $A$ and which is both open and closed in $A$. Then, clearly, the sets $U$ and $U^c$ (both open in $A$ ) disconnect $A$. Conversely, suppose that $A$ is disconnected by sets $U$ and $V$ (both open in $A$ ). Then the set $U$ is not the null set, is not equal to $A$ (because $V$, its complement with respect to $A$, is nonempty), is open in $A$, and is closed in $A$ (because $V$ is open in $A$ ).
