A.1 ^{1} Give an example of a metric space (E, d) and infinitely many open sets {U_{i} ⊂ E}_{i∈I} such that ⋂_{i∈I} U_{i} is not open. You could consider suitable intervals in ℝ for example.

Proof. Take E to be ℝ, under the standard metric d(x, y) = |x − y| for x, y ∈ E. Then, let U_{i} be the open interval (1 − \frac{1}{n}, 1 + \frac{1}{n}). Then, as i → ∞, (1 − \frac{1}{n}, 1 + \frac{1}{n}) → [1, 1], which is clearly not open in ℝ. □

A.2 Let X be a subset of a metric space (E, d). Show that:

\overline{X}^{∘}=\overline{\overline{X}^{∘}}^{∘} and that \overline{X^{∘}} = \overline{\overline{X^{∘}}^{∘}}.

Give an example of a subset X of ℝ such that the following sets:

X, \overline{X}, X^{∘}, \overline{X}^{∘}, \overline{X^{∘}}, \overline{\overline{X}^{∘}}, \overline{X^{∘}}^{∘},

are pairwise distinct.

Proof of Part 1. Let α ∈\overline{X}^{∘}. Then, for all F ⊂ E, where F ⊃\overline{X}^{∘}, α ∈ F. Thus, α ∈ \overline{\overline{X}^{∘}}. Since

α was arbitrary, \overline{X}^{∘}⊂ \overline{\overline{X}^{∘}}. Thus, since \overline{X}^{∘} is open and a subset of \overline{\overline{X}^{∘}}, it is also a subset of \overline{\overline{X}^{∘}}^{∘}

(which, by definition, is the union of all open subsets of \overline{\overline{X}^{∘}}).

The interior of \overline{X} is the same as the complement of the closure of the complement of \overline{X}

(in fact, this is true for all subsets in general). Thus, \overline{X}^{∘}= \overline{\overline{X}^{C}}^{C} and \overline{\overline{X}^{∘}}^{∘}= \overline{\overline{\overline{\overline{X}^{C}}^{C}}^{C}}^{C}. Now,

\overline{X}^{C} ⊂ \overline{\overline{X}^{C}}, by definition, so \overline{X} ⊃ \overline{\overline{X}^{C}}^{C}, and thus \overline{X} ⊃ \overline{\overline{\overline{X}^{C}}^{C}}. Thus,

⟹ \overline{X}^{C} ⊂ \overline{\overline{\overline{X}^{C}}^{C}}^{C}

⟹ \overline{\overline{X}^{C}} ⊂ \overline{\overline{\overline{\overline{X}^{C}}^{C}}^{C}}
⟹ \overline{\overline{X}^{C}}^{C} ⊃ \overline{\overline{\overline{\overline{X}^{C}}^{C}}^{C}}^{C}

Thus, \overline{X}^{∘}⊃\overline{\overline{X}^{∘}}^{∘}. ∴\overline{X}^{∘}=\overline{\overline{X}^{∘}}^{∘} □

Part Two. We have

\overline{\overline{X}^{C}}^{C} = \overline{\overline{\overline{\overline{X}^{C}}^{C}}^{C}}^{C}

1Many thanks to Eric Suh for making his L^{A}TEXfile available to my disposal.

for any X ⊂ (E, d), and we want to show that

\overline{X^{∘}} = \overline{\overline{X^{C}}^{C}} = \overline{\overline{\overline{\overline{X^{C}}^{C}}^{C}}^{C}} = \overline{\overline{X^{∘}}^{∘}}

We know that

\overline{\overline{(X^{C})}^{C}}^{C} = \overline{\overline{\overline{\overline{(X^{C})}^{C}}^{C}}^{C}}^{C}

since X^{C} ⊂ (E, d). Therefore,

\overline{\overline{X^{C}}^{C}} = \overline{\overline{\overline{\overline{X^{C}}^{C}}^{C}}^{C}}

And thus

\overline{X^{∘}} = \overline{\overline{X^{∘}}^{∘}}

□

Example Set. Let X = (1, 3]\{2} ∪ ([4, 5] ∩ ℚ) ∪ {0} ⊂ ℝ. \overline{X} = [1, 3] ∪ [4, 5] ∪ \{0\}

X^{∘}= (1, 2) ∪ (2, 3)

\overline{X}^{∘}= (1, 3) ∪ (4, 5)

\overline{X^{∘}} = [1, 3]

\overline{\overline{X}^{∘}} = [1, 3] ∪ [4, 5]

\overline{X^{∘}}^{∘}= (1, 3) □

A.3 [Induced Topology] Let (E, d) be a metric space, and let X be a subset of E so that (X, d) is itself a topological space with the induced distance. Show that U ⊂ X is open in X if and only if there exists an open set V of E such that U = X ∩ V. Be careful: if U ⊂ X is open in X, it is not necessarily open in E.

Proof. Let U ⊂ X be open in X. For every point x ∈ U, there exists an open ball of radius r_{x}: B_{x}(x, r_{x}) ⊂ U. Now, consider the same open balls, but consider them as open balls in E instead of X. Then, let V be the union of all such open balls. V is open in E, and U = X ∩V.

Now going in the other direction, let V be an open set in E such that U = X ∩ V. For all x ∈ U, there exists B_{E}(x, r) ⊂ V. Construct B_{X}(x, r) = B_{E}(x, r) ∩ X. This is an open ball in X, and since this exists for all x ∈ U, U is open in X. □

A.4 [Product Topology] Let (E_{1}, d_{1}) and (E_{2}, d_{2}) be two metric spaces. Define a distance d on

E_{1} × E_{2} by

d((x_{1}, x_{2}), (y_{1}, y_{2})) = d_{1}(x_{1}, y_{1}) + d_{2}(x_{2}, y_{2}).

1. Show that U ⊂ E_{1} × E_{2} is open if and only if for each point (u_{1}, u_{2}) ∈ U, there exists U_{1} and U_{2} open in E_{1} and E_{2} with (u_{1}, u_{2}) ∈ U_{1} × U_{2} and U_{1} × U_{2} ⊂ U.

2. If X is another metric space, show that (f_{1}, f_{2}) : X → E_{1} × E_{2} is continuous if and only if f_{1} and f_{2} both are.

Proof of Part 1. Let U ⊂ E_{1} ×E_{2} be open, and let (u_{1}, u_{2}) ∈ U. Since U is open, there exists r > 0 such that B ((u_{1}, u_{2}), r) ⊂ U. Thus, for any (v_{1}, v_{2}) in the open ball around (u_{1}, u_{2}),

d((u_{1}, u_{2}), (v_{1}, v_{2})) = d_{1}(u_{1}, v_{1}) + d_{2}(u_{2}, v_{2}) < r.

Clearly, we see that this implies d_{1}(u_{1}, v_{1}) < r and d_{2}(u_{2}, v_{2}) < r. Now let U_{1} be the open ball in E_{1} around u_{1} with radius r/2 and U_{2} be the open ball in E_{2} around u_{2} with radius r/2. By definition, U_{1} and U_{2} are open. U_{1} × U_{2} ⊂ U, since anything within the cartesian product must lie within the open ball in U around (u_{1}, u_{2}).

Now, assume that for each point (u_{1}, u_{2}) ∈ U there exists U_{1} and U_{2} open in E_{1} and E_{2} with (u_{1}, u_{2}) ∈ U_{1} × U_{2} and U_{1} × U_{2} ⊂ U. Then there is an open ball in U_{1} around u_{1} with radius r_{1}, and an open ball in U_{2} around u_{2} with radius r_{2}. Let r = min(r_{1}, r_{2}). Then, for any element (v_{1}, v_{2}) ∈ U such that d ((u_{1}, u_{2}), (v_{1}, v_{2})) < r is within U_{1} × U_{2}. Thus, the open ball resides completely in U, and since such an open ball exists for every point (u_{1}, u_{2}) in U, U is open. □

Proof of Part 2. For all x ∈ X, and for all ε > 0, there is a δ > 0 such that for all x′ ∈ X, d(x, x′) < δ ⟹ d_{1}(f_{1}(x), f_{1}(x′)) + d_{2}(f_{2}(x), f_{2}(x′)) < ε. Thus, for all x′ ∈ X,

d(x, x′) < δ ⟹ d_{1}(f_{1}(x), f_{1}(x)′) < ε

and

d(x, x′) < δ ⟹ d_{2}(f_{2}(x), f_{2}(x)′) < ε

For all x ∈ X and for all ε > 0, we must find a δ > 0 such that for all x′ ∈ X, d(x, x′) < δ ⟹ d_{1}(f_{1}(x), f_{1}(x)′)+d_{2}(f_{2}(x), f_{2}(x)′) < ε. Since f_{1} and f_{2} are both continuous, for some ε/2, there is a δ_{1} > 0 such that d_{1}(f_{1}(x), f_{1}(x)′) < ε/2 if d(x, x′) < δ_{1}, and for some ε/2, there is a δ_{2} > 0 such that d_{2}(f_{2}(x), f_{2}(x)′) < ε/2 if d(x, x′) < δ_{2}. Now, let δ = min(δ_{1}, δ_{2}). Then,

d(x, x′) < δ ⟹ d_{1}(f_{1}(x), f_{1}(x′)) + d_{2}(f_{2}(x), f_{2}(x′)) < ε/2 + ε/2 = ε

□