Topology

Suppose a neighbourhood \(O_{r}(p)\) contains finitely many points of \(E\) except possibly \(p\) itself, \(\{ q_1, \dots , q_{n} \}\). Take \(\hat{r} = \min_{i} d(q_{i}, p) > 0\). Then, \((O_{\hat{r}}(p) \setminus \{ p \}) \cap E = \emptyset\), which contradicts with \(p\) being a limit point.

Let \(F\) be a set and \(E\) be its complement.

Assume \(F\) is closed. Let \(x \in E\), \(x \notin F\). Then \(x\) cannot be a limit point of \(F\). By definition, there exists a neighbourhood \(O\) of \(x\) s.t. \((O \setminus \{ x \}) \cap F = O \cap F = \emptyset\). Thus, \(O \subset E\). \(E\) is open.

Conversely, assume \(E\) is open. Let \(x\) be a limit point of \(F\). Then every neighbourhood of \(x\) intersects with \(F\) at some point other than \(x\). Thus, neighbourhoods of \(x\) cannot be completely inside \(E\) and hence \(x\) cannot be an interior point of \(E\). Since \(E\) is open, this means that \(x \notin E\), i.e., \(x \in F\).

Let’s rewrite the condition for \(x\) being a limit point of \(A \subset X\): \(\forall \epsilon > 0, (O_{\epsilon}(x) \cap A) \setminus \{ x \} \neq \emptyset\). This is the same as \(\forall \epsilon > 0, \exists y \in A, y \neq x, d(x, y) < \epsilon\). It is easy to verify that this is equivalent to the statement in the proposition.

This means that if \(K, F\) are disjoint then they are actually separated by some distance.

Suppose the opposite. We can take two sequence \(\{ p_{n} \} \subset K, \{ q_{n} \} \subset F\) that can be arbitrarily close. By Thm. 1, \(\{ p_{n} \}\) has some subsequence \(\{ p_{n_{k}} \}\) that converges to some \(p \in K\). Then, \(\{ q_{n_{k}} \}\) also converges to \(p\). Since \(F\) is closed, \(p\) is in \(F\), which contradicts with the condition that \(F \cap K = \emptyset\).

Since \(B\) is bounded below, \(L\) is non-empty. By the lub property of \(C\), \(L\) has a least upper bound. Let us denote it by \(\alpha\). Since every element in \(B\) is an upper bound of \(L\), \(\alpha \preceq\) all of them. Then, by the definition of \(L\), \(\alpha \in L\). Then, \(\alpha\) is the greatest among \(L\), i.e., the greatest lower bound of \(B\).

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Triangle inequality.

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The sequence formulation of function limit implies that the existence of some convergent sequence \(f(x_{n})\). By the uniqueness of sequence limit, the function limit is also unique.

From the arithmetics rules for sequential limits.

Let \(p\) be an isolated point of \(E\). Pick \(\delta > 0\) s.t. the only \(x \in E\) for which \(d_{X}(x, p) < \delta\) is \(p\). Then \(f\) is continuous at \(p\).

Do the epsilon-delta twice. Let \(\epsilon > 0\). \(g\) is continuous at \(f(p)\), so \(\exists \eta > 0\) s.t. \(d_{Y}(y, f(p)) < \eta \implies d_{E}(g(y), g(f(p))) < \epsilon\). \(f\) is continuous at \(p\), so \(\exists \delta > 0\) s.t. \(d_{X}(x, p) < \delta \implies d_{Y}(f(x), f(p)) < \eta\). This means that \(d_{X}(x, p) < \delta \implies d_{E}(h(x), h(p)) < \epsilon\).

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\(y \in f(\cl{A}) \iff \exists x \in \cl{A}, y = f(x) \iff \exists \{ x_{n} \} \subset A, x_{n} \to x, y = f(x)\).

By the continuity of \(f\), this implies \(\exists \{ x_{n} \} \subset A, f(x_{n}) \to f(x) = y\). Thus, \(\exists \{ y_{n} \} \subset f(A), y_{n} \to y\). This is exactly the condition for \(y\) being in \(\cl{f(A)}\).

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Let \(\epsilon > 0\). We want to find a uniform \(\delta > 0\) s.t. \(d_{X}(p, q) < \delta \implies d_{Y}(f(p), f(q)) < \epsilon\). By the continuity, for every \(p \in X\), \(\exists \delta_{p} > 0\) s.t. \(d_{X}(p, q) < \delta_{p} \implies d_{Y}(f(p), f(q)) < \epsilon\). Let \(J(p) := \{ q \in X: d_{X}(p, q) < \frac{1}{2} \delta_{p} \}\). Then \(\cup _{p \in X} J(p)\) is an open cover of \(X\). By the compactness of \(X\), there exists a finite sub collection \(\cup _{i=1} ^{n} J(p_{i})\) which is also an open cover of \(X\). Take \(\delta = \frac{1}{2} \min _{i=1} ^{n} \{ \phi(p_{i}) \} > 0\), where the min makes sense and is \(> 0\) because the collection is finite. Then, for any \(p, q\) with \(d_{X}(p, q) < \delta\), \(p\) must be in some \(J(p_{i})\), and hence both \(p, q\) are within a distance of \(\delta_{p_{i}})\) to \(p_{i}\), implying that \(d_{Y}(p, q) < d_{Y}(p, p_{i}) + d_{Y}(p_{i}, q) = \epsilon\).

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For the second one, any topology \(\mathcal{T} '\) containing all \(\mathcal{T}_{\alpha}\) would have to contain all finite intersections of elements of \(\cup _{\alpha} \mathcal{T}_{\alpha}\). The collection of all finite intersections is the basis \(\mathcal{B}\) corresponding to the subbasis \(\mathcal{S}\). Consequently, \(\mathcal{T}'\) is finer than the topology generated by \(\mathcal{B}\).

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That is equivalent to \(A'_{Y} = A'_{X} \cap Y\) (the set of limit points in \(Y\), the set of limit points in \(X\)).

By Thm. 1, \(x \in A'_{Y}\) iff \(\forall B \in \mathcal{B}_{Y}, x \in B \implies B \cap A \setminus \{ x \} \neq \emptyset\). By Prop. 1, this means that \(\forall B \in \mathcal{B}_{X}, x \in B \cap Y \implies B \cap Y \cap A \setminus \{ x \} \neq \emptyset\), which is exactly the condition for \(x \in A'_{X} \cap Y\).

1. Let \(V\) be open in \(Y\). If \(V \ni x\), \(f ^{-1}(x) = X\); otherwise, \(f ^{-1}(x) = \emptyset\).
2. Let \(V\) be open in \(X\). Then \(j ^{-1}(V) = V \cap A\), which is open.
3. Let \(V\) be open in \(Z\). Then \(g ^{-1}(U)\) is open in \(Y\) and \(f ^{-1}(g ^{-1}(U))\) is open in \(X\). \((g \circ f) ^{-1}(U) = f ^{-1}(g ^{-1}(U))\).
4. \(f_{|A}\) is the composition of the inclusion map \(j: A \to X\) and \(f\).

5. If \(f(X) \subset Z \subset Y\): Let \(V\) be open in \(Z\). Then \(V = Z \cap U\) for some open set \(U\) of \(Y\). Since \(f ^{-1}\) and \(g ^{-1}\) coincide on \(f(X)\), \(g ^{-1}(V) = g ^{-1}(V \cap f(X)) = f ^{-1}(V \cap f(X)) = f ^{-1}(U \cap f(X)) = f ^{-1}(U)\), which is open.

   If \(Y \subset Z\): Then \(g\) is the composition of the inclusion map \(j: Y \to Z\) and \(f\).

6. Let \(V\) be open in \(Y\). Then \(f ^{-1}(V) \cap U_{\alpha} = (f|_{U_{\alpha}}) ^{-1}(V)\) for all \(\alpha\). Thus, \(f ^{-1}(V) = \cup _{\alpha}(f ^{-1}(V) \cap U_{\alpha})\) is open.

Use the definition of contiuity by closed sets.

Let \(\beta \in J\). For every open set \(U_{\beta} \in X_{\beta}\), \(\pi_{\beta} ^{-1}(U_{\beta}) = \prod_{\alpha \in J} U_{\alpha}\), where \(U_{\alpha} = X_{\alpha}\) for \(\alpha \neq \beta\). This is an open set in either the box topology or the product topology.

Let \(\pi_{\beta}\) be the projection map (Defn. 1) of the product onto its \(\beta\)-th coordinate.

Assume each function \(f\) is continuous. Then \(f_{\beta} = \pi_{\beta} \circ f\) is continuous, as it is the composition of two continuous functions.

Conversely, assume each coordinate function \(f_{\alpha}\) is continuous. By Thm. 1, it suffices to show that the pre-image of each subbasis element is open in \(A\). We take the typical subbasis in the product topology: \(\{ \pi_{\beta} ^{-1}(U_{\beta}) | \beta \in J, U_{\beta} \text{ open in } X_{\beta} \}\). Then \(f ^{-1}(\pi_{\beta} ^{-1}(U_{\beta})) = f _{\beta} ^{-1}(U_{\beta})\) (only has requirements on the \(\beta\)-th coordinate). This set is open in \(A\) because \(f_{\beta} ^{-1}\) is continuous.

Summary: We make use the subbasis elements, whose pre-images are only about one coordinate.

The first direction can be proved in the same way.

Now, assume each \(f_{\alpha}\) is continuous and try to show that the pre-image of each basis element is open in \(A\) (Thm. 1)). Consider a typical basis element of the box topology, \(B = \prod_{\beta \in J} U_{\beta}\), where \(U_{\beta}\) is open in \(X_{\beta}\). Then,

This is where problems may arise: By the continuity of each \(f_{\beta}\), each \(f_{\beta} ^{-1}(U_{\beta})\) is open in \(A\). However, there is no guarantee that (countably) infinite intersection of open sets is still open.

To find a counterexample, we just need to construct a countable infinite intersection of open sets that is not open. E.g., in \(\mathbb{R} \), we have \(\cap _{i=1} ^{\infty} (-1 /n, -1 /n) = \{ 0 \}\), \(\cap _{i=1} ^{\infty} (0, 1 + 1 /n) = (0, 1]\), etc. And we can choose \(f_{\beta}\) to be simply the identity map. All together, here is a complete example: \(f: \mathbb{R} \to \mathbb{R} ^{\omega}\), where \(\mathbb{R} ^{\omega} = \prod_{n=1} ^{\infty} \mathbb{R} \). \(f(x) = (f_{n}(x))_{n \in \mathbb{Z} _{+}}\), where \(f_{n}(x) = x\). \(B = \prod_{n=1} ^{\infty} (-1 /n, 1 /n)\).