Technology and Art
This post lists solutions to the exercises in the Compactness and Finite Dimension section 2.5 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.
Proof:
\(\mathbb{R}^n\) has a sequence \(x_n=1,2,3,4, \cdots\) which does not have a convergent subsequence, since every \(d(x_m,x_n) \geq 1\) for all \(m,n \in \mathbb{N}\). The sequence exists in \(\mathbb{C}^n\), implying that every sequence in that space does not have a convergent subsequence.
Thus \(\mathbb{R}^n\) and \(\mathbb{C}^n\) are not compact.
\[\blacksquare\]Proof:
The discrete metric space \(X\) has the distance metric as:
\[d(x,y)=\begin{cases} 0 & \text{if } x=y \\ 1 & \text{if } x \neq y \end{cases}\]Assume any sequence \((x_n) \subset X\) which does not repeat its elements. Then \(d(x_m,x_n)=1\) for all \(m,n: m \neq n, m,n \in \mathbb{N}\). Thus, this sequence has no convergent subsequence. Thus the discrete metric space \(X\) is not compact.
\[\blacksquare\]Answer:
\(y=x\) is not compact because it is not bounded.
\(y=\sin x, x \in [0, 2\pi]\) is compact because \([0,2\pi]\) is compact and sine is a continuous function, and we know that continuous functions map compact sets to compact sets.
Proof:
The metric of \(s\) is defined by:
\[d(x,y) = \sum\limits_{j=1}^\infty \frac{1}{2^j} \frac{|\xi_j-\eta_j|}{1+|\xi_j-\eta_j|}\]\(\xi_k(x)\) extracts the \(k\)-th element of the sequence \(x\).
Assume a sequence \((p_n) \subset M\), like so:
\[p_1 = x^1_1, x^1_2, x^1_3, \cdots, x^1_k, \cdots \\ p_2 = x^2_1, x^2_2, x^2_3, \cdots, x^2_k, \cdots \\ \vdots \\ p_m = x^m_1, x^m_2, x^m_3, \cdots, x^m_k, \cdots \\ \vdots\]For fixed \(k\), we have \(\vert \xi_k(p_m) \vert < \gamma_k\). Thus \(x^m_k\) is bounded for fixed \(k\).
Set \(k=1\), then \(x^1_1, x^2_1, x^3_1, \cdots\) is bounded. Thus, by the Bolzano-Weierstrauss Theorem, this contains a convergent subsequence. Let this convergent subsequence converge to \(x_1\). Let \(P_1\) be the set of sequences which contain these subsequence entries. We can repeat the same exercise for \(k=2\) and apply it to \(P_1\) to get \(P_2\), etc.
We finally get a subsequence of \((p_n)\) (call it \(p_{n_j}\)) where, for any given \(k\), we have a convergent sequence \((x^{n_j}_k)\) converging to \(x_k\). We construct a sequence out of these limits \(p=x_1, x_2, \cdots\). For any \(p_m\), we have:
\[d(p_m, p) = \sum\limits_{j=1}^\infty \frac{1}{2^j} \frac{|x^m_j-x_j|}{1+|x^m_j-x_j|}=\sum\limits_{j=1}^\infty \frac{1}{2^j} \left( 1 - \frac{1}{1+|x^m_j-x_j|} \right)\]Take \(\epsilon > 0\). Assume \(N_1 \in \mathbb{N}\) such that \(\vert x^m_1-x_1 \vert\), similarly for \(N_2\), and so on. Now take \(N=\max{(N_1, N_2, \cdots)}\). Then for \(m>N\), we have:
\[d(p_m, p) = \sum\limits_{j=1}^\infty \frac{1}{2^j} \left( 1 - \frac{1}{1+\epsilon} \right) \\ \lim_{\epsilon \rightarrow 0} d(p_m, p) = 0\]Thus, \(p\) is the limit of the constructed subsequence \(p_{n_j}\). Thus \((p_n)\) has a convergent subsequence. Since \((p_n)\) was arbitrary, \(M\) is compact.
\[\blacksquare\]Proof:
Take \(x \in \mathbb{R}\). Then, for any \(\epsilon>0\), the set \([x-\epsilon, x+\epsilon]\) is closed and bounded and contains the \(\epsilon\)-neighbourhood, and is thus a compact neighbourhood of \(x\). Thus \(\mathbb{R}\) is locally compact.
Take \(x \in \mathbb{R}^n\). For any \(\epsilon>0\), the closed ball \(\bar{B}(x, \epsilon)\) contains the open \(\epsilon\)-neighbourhood \(B(x,\epsilon)\) and is thus a compact neighbourhood of \(x\). Thus \(\mathbb{R}^n\) is locally compact.
\[\blacksquare\]Proof:
The neighbourhood of a point \(x\) is defined as a set which contains an open ball centered on \(x\) (or open set containing \(x\)).
Since any open ball around \(x \in X\) is also contained in the set \(X\), \(X\) is a neighbourhood of \(x\). Since \(X\) is compact, \(x\) has a compact neighbourhood, implying that \(X\) is locally compact.
\[\blacksquare\]Proof:
(TODO)
\[\blacksquare\]Proof:
Norms are continuous. If \(f(x)=\frac{\|x\|}{ {\|x\|}_2}\) is also continuous, assuming that \({\|x\|}_2 \neq 0\). Assume a compact subset \(M \subset X\); then \(f(M)\) is also compact, and thus contains a maximum and a minimum value. Specifically, it contains a infimum, call it \(a\). Then we have:
\[f(x) = \frac{\|x\|}{ {\|x\|}_2} \geq a \\ {\|x\|} \geq a {\|x\|}_2\]If \({\|x\|}_2 =0\), then obviously \({\|x\|} \geq a {\|x\|}_2 = 0\).
\[\blacksquare\]Proof:
Every sequence in \(X\) has a convergent subsequence.
Choose an arbitrary sequence \((x_n) \subset M\). Since \((x_n) \subset X\), it has a convergent subsequence \((x_{n_k}\), therefore this subsequence has a limit point \(x\).
We note that \((x_{n_k}) \subset M\).
Since \(M\) is closed, it contains all its limit points, thus \(x \in M\). Thus \(M\) contains the limit of this subsequence as well. Thus the sequence \((x_n)\) has a convergent subsequence in \(M\). Since \((x_n)\) is arbitrary, all sequences in \(M\) have convergent subsequences. Thus, \(M\) is compact.
\[\blacksquare\]Proof:
For \(T\) to be a homeomorphism, its inverse should be continuous.
\(X\) is compact and \(T\) is continuous, thus \(T(X)\) is compact. Choose any sequence \((x_n) \subset X\). Since \(X\) is compact, \((x_n)\) contains a convergent subsequence \((x_{n_k})\) which converges to \(x\). Since continuous functions map convergent sequences to convergent sequences, \((y_{n_k}) = T((x_{n_k}))\) is a convergent sequence.
Now we define \(T^{-1}\) as \(T^{-1}: (y_{n_k}) \mapsto (x_{n_k})\). We know that if a function is continuous if and only if it maps convergent sequences to convergent sequences.
Since both \((x_{n_k})\) and \((y_{n_k})\) are convergent sequences, \(T^{-1}\) is continuous at \(x\).
Since \((x_n)\) was arbitrarily chosen, \(T\) is a homeomorphism.
\[\blacksquare\]