Sample Midsem Solution

(a) Use the Cauchy definition with \(\varepsilon=1\). There is \(N\) such that \(|x_n-x_m|<1\) for \(m,n\ge N\). Taking \(m=N\), we get \(|x_n|<|x_N|+1\) for \(n\ge N\). Therefore \[M=\max\{|x_1|,\dots,|x_{N-1}|,|x_N|+1\}\] bounds every term.

(b) If \(m>n\), then \[|x_m-x_n|\le\sum_{k=n}^{m-1}|x_{k+1}-x_k|\le\sum_{k=n}^{m-1}2^{-k}<2^{-n+1}.\] Given \(\varepsilon>0\), choose \(N\) with \(2^{-N+1}<\varepsilon\). Then \(m>n\ge N\) implies \(|x_m-x_n|<\varepsilon\). Thus \((x_n)\) is Cauchy and converges in \(\mathbb{R}\).

Supremum. For all \(m,n\), \(\frac1m-\frac1n<\frac1m\le1\), so \(1\) is an upper bound and is never attained. Given \(\varepsilon>0\), choose \(n\) with \(1/n<\varepsilon\). Taking \(m=1\) gives \(1-1/n\in A\) and \(1-1/n>1-\varepsilon\). Hence \(\sup A=1\).

Infimum. Swapping \(m\) and \(n\) negates an element, so \(A=-A\). Therefore \(\inf A=-\sup A=-1\). Neither bound is attained.

(a) Let \(F=\bigcap_\alpha F_\alpha\), with every \(F_\alpha\) closed. De Morgan's law gives \(F^c=\bigcup_\alpha F_\alpha^c\). Each complement is open, and an arbitrary union of open sets is open. Hence \(F^c\) is open and \(F\) is closed.

(b) It is not compact. Take \(x=1/\sqrt2\in[0,1]\setminus\mathbb Q\). Every neighbourhood of \(x\) contains rationals by density of \(\mathbb Q\), so \(x\) is a limit point of \(\mathbb Q\cap[0,1]\) not contained in the set. The set is not closed and, by Heine–Borel, not compact.

Use sequential compactness. Let \((x_n)\) be any sequence in \(K\cap F\). Compactness of \(K\) gives a subsequence \(x_{n_k}\to x\) with \(x\in K\). Every subsequence term lies in the closed set \(F\), so its limit lies in \(F\). Thus \(x\in K\cap F\). Every sequence in the intersection has a subsequence converging to a point of the intersection, so \(K\cap F\) is sequentially compact and hence compact.

(a) A continuous image of the compact domain \([0,1]\) must be compact. But \((0,1)\) is not closed and hence is not compact. Therefore it cannot be the image.

(b) A continuous image of the connected interval \([0,1]\) must be connected, hence an interval in \(\mathbb R\). But \([0,1]\cup[2,3]\) contains \(1\) and \(2\) but not \(3/2\), so it is not an interval. Therefore it cannot be the image.

Failure on \([1,\infty)\). Set \(\varepsilon=1\). For any \(0<\delta<1\), choose \(x=1/\delta\) and \(y=1/\delta+\delta/2\). Then \(|x-y|=\delta/2<\delta\), but \[|f(y)-f(x)|=\left|\left(\frac1\delta+\frac\delta2\right)^2-\frac1{\delta^2}\right|=1+\frac{\delta^2}{4}\ge1.\] Thus uniform continuity fails.

Success on \([1,2]\). The function is continuous and the interval is compact. By Heine–Cantor, \(f\) is uniformly continuous on \([1,2]\).

For \(x>0\), \(f(x)=x^2\), so \(f'(x)=2x\). For \(x<0\), \(f(x)=-x^2\), so \(f'(x)=-2x\). At zero, \[\frac{f(h)-f(0)}h=\frac{h|h|}{h}=|h|\longrightarrow0,\] so \(f'(0)=0\). Hence \(\boxed{f'(x)=2|x|}\) for every \(x\).

However, \(\frac{f'(h)-f'(0)}h=\frac{2|h|}{h}\) equals \(2\) for \(h>0\) and \(-2\) for \(h<0\). The two-sided limit does not exist, so \(f'\) is not differentiable at \(0\).

Existence. The polynomial is continuous, \(p(-1)=-3<0\), and \(p(0)=1>0\). By the Intermediate Value Theorem, some \(c\in(-1,0)\) satisfies \(p(c)=0\).

Uniqueness. If distinct roots \(a<b\) existed, Rolle's theorem would give \(c\in(a,b)\) with \(p'(c)=0\). But \(p'(x)=5x^4+3\ge3>0\) for every real x, a contradiction. Hence exactly one real root exists.