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Week 5 Discussion Notes

Table of Contents

Homework Comments

To prove the triangle inequality for the ℓ1\ell^1 metric, you needed to use a limit law:

∑n=1∞∣an−bn∣=lim⁡N→∞∑n=1N∣an−bn∣≤lim⁡N→∞(∑n=1N∣an−cn∣+∑n=1N∣cn−bn∣)=lim⁡N→∞∑n=1N∣an−cn∣+lim⁡N→∞∑n=1N∣cn−bn∣(limit law; both sums converge)=∑n=1∞∣an−cn∣+∑n=1∞∣cn−bn∣.\begin{aligned} \sum_{n=1}^\infty \abs{a_n - b_n} &= \lim_{N\to\infty} \sum_{n=1}^N \abs{a_n - b_n} \\ &\leq \lim_{N\to\infty} \p{\sum_{n=1}^N \abs{a_n - c_n} + \sum_{n=1}^N \abs{c_n - b_n}} \\ &= \lim_{N\to\infty} \sum_{n=1}^N \abs{a_n - c_n} + \lim_{N\to\infty} \sum_{n=1}^N \abs{c_n - b_n} && \p{\text{limit law; both sums converge}} \\ &= \sum_{n=1}^\infty \abs{a_n - c_n} + \sum_{n=1}^\infty \abs{c_n - b_n}. \end{aligned}

Also, to show that the unit ball

B={{an}n∈ℓ1 | ∑n=1∞∣an∣≤1}B = \set{\set{a_n}_n \in \ell^1 \st \sum_{n=1}^\infty \abs{a_n} \leq 1}

is closed and bounded, my hint was to try to get you to recognize this set as

B={x∈ℓ1 | d(x,0)≤1},B = \set{x \in \ell^1 \st d\p{x, 0} \leq 1},

i.e., as a closed ball. We proved in lecture that closed balls are closed, and this set is bounded because B⊆B2(0)B \subseteq B_2\p{0} (remember that our definition of bounded involves containing the set in an open ball).

Continuity

Definition

A function f ⁣:(X,dX)→(Y,dY)\func{f}{\p{X, d_X}}{\p{Y, d_Y}} is continuous at x0∈Xx_0 \in X if for for every ε>0\epsilon > 0, there exists δ>0\delta > 0 such that

dX(x,x0)<δ  ⟹  dY(f(x),f(x0))<ε.d_X\p{x, x_0} < \delta \implies d_Y\p{f\p{x}, f\p{x_0}} < \epsilon.

If ff is continuous at every x0∈Xx_0 \in X, then we say that ff is continuous.
Theorem

Let f ⁣:(X,dX)→(Y,dY)\func{f}{\p{X, d_X}}{\p{Y, d_Y}} be a function. The following are equivalent:

  1. ff is continuous.
  2. For any x∈Xx \in X and any sequence {xn}n\set{x_n}_n in XX such that xn→dXxx_n \xrightarrow{d_X} x, we have f(xn)→dYf(x)f\p{x_n} \xrightarrow{d_Y} f\p{x}.
  3. For any open set V⊆YV \subseteq Y, the preimage f−1(V)f^{-1}\p{V} is open.
  4. For any closed set F⊆YF \subseteq Y, the preimage f−1(F)f^{-1}\p{F} is closed.
Remark.

Note that (iii) doesn't refer to the metric at all, so in more advanced math classes, (iii) is usually taken to be the definition of continuity.

We like continuity because it lets us move limits around. For example, we can rewrite (ii) as

lim⁡n→∞f(xn)=f(lim⁡n→∞xn).\lim_{n\to\infty} f\p{x_n} = f\p{\lim_{n\to\infty} x_n}.

Continuity is one of the biggest reasons why we're studying metric spaces to begin with. Before, we only had continuity defined for functions involving real numbers, but there are plenty of interesting functions defined on more general metric spaces. For example, the Fourier transform is defined on function spaces, and we can now talk about whether it's continuous or not.

Now that we have the notion of continuity, a reasonable question to ask if whether every metric space has a (non-trivial) continuous function or not. The answer is yes:

Example 1.

Let (X,d)\p{X, d} be a (non-empty) metric space. For a fixed y∈Xy \in X, define f ⁣:X→R\func{f}{X}{\R} via f(x)=d(x,y)f\p{x} = d\p{x, y}. Show that ff is continuous.
Solution.

For this problem, we'll want to work directly from the definition. Let ε>0\epsilon > 0. Like usual, we're going to work backwards to find δ\delta:

∣f(x)−f(x0)∣=∣d(x,y)−d(x,x0)∣≤d(x,x0)\abs{f\p{x} - f\p{x_0}} = \abs{d\p{x, y} - d\p{x, x_0}} \leq d\p{x, x_0}

by the reverse triangle inequality. So, if we let δ=ε\delta = \epsilon, then whenever d(x,x0)<δd\p{x, x_0} < \delta, we have

∣f(x)−f(x0)∣≤d(x,x0)<δ=ε,\abs{f\p{x} - f\p{x_0}} \leq d\p{x, x_0} < \delta = \epsilon,

so ff is continuous.
Example 2.

Let (X,d)\p{X, d} be a metric space with the discrete metric and (Y,d)\p{Y, d} be any metric space. Then every function f ⁣:X→Y\func{f}{X}{Y} is continuous.
Solution.

For this problem, (iii) is the easiest to prove: given any open set V⊆YV \subseteq Y, the preimage f−1(V)f^{-1}\p{V} is automatically open because every set is open with the discrete metric.
Example 3.

Let

X={{an}n∈ℓ∞ | lim⁡n→∞an exists}d({an}n,{bn}n)=sup⁡n∣an−bn∣,\begin{gathered} X = \set{\set{a_n}_n \in \ell^\infty \st \lim_{n\to\infty} a_n \text{ exists}} \\ d\p{\set{a_n}_n, \set{b_n}_n} = \sup_n{\abs{a_n - b_n}}, \end{gathered}

i.e., XX is the set of bounded convergent sequences. Define f ⁣:X→R\func{f}{X}{\R} via

f({an}n)=lim⁡n→∞an.f\p{\set{a_n}_n} = \lim_{n\to\infty} a_n.

Show that ff is continuous.
Solution.

For this problem, (iii) and (iv) aren't that useful just because it's hard to make sense of f−1(V)f^{-1}\p{V}. So, we'll want to use an ε\epsilon-δ\delta proof.

Let ε>0\epsilon > 0. Working backwards,

∣f({an}n)−f({bn}n)∣=∣lim⁡n→∞an−lim⁡n→∞bn∣=∣lim⁡n→∞(an−bn)∣(limit law)=lim⁡n→∞∣an−bn∣(∣ ⋅ ∣ is continuous).\begin{aligned} \abs{f\p{\set{a_n}_n} - f\p{\set{b_n}_n}} &= \abs{\lim_{n\to\infty} a_n - \lim_{n\to\infty} b_n} \\ &= \abs{\lim_{n\to\infty} \p{a_n - b_n}} && \p{\text{limit law}} \\ &= \lim_{n\to\infty} \abs{a_n - b_n} && \p{\abs{\:\cdot\:}\text{ is continuous}}. \end{aligned}

Notice that by definition,

∣an−bn∣≤sup⁡n∣an−bn∣=d({an}n−{bn}n),\abs{a_n - b_n} \leq \sup_n{\abs{a_n - b_n}} = d\p{\set{a_n}_n - \set{b_n}_n},

so we get

∣f({an}n)−f({bn}n)∣≤d({an}n−{bn}n).\abs{f\p{\set{a_n}_n} - f\p{\set{b_n}_n}} \leq d\p{\set{a_n}_n - \set{b_n}_n}.

Thus, setting δ=ε\delta = \epsilon works.

Compactness and Continuity

Proposition

Let f ⁣:(X,dX)→(Y,dY)\func{f}{\p{X, d_X}}{\p{Y, d_Y}} be continuous. Then for any compact set K⊆XK \subseteq X, its image f(K)⊆Yf\p{K} \subseteq Y is compact.

Usually, we say "the continuous image of a compact set is compact."

Example 4.

If f ⁣:(X,dX)→(Y,dY)\func{f}{\p{X, d_X}}{\p{Y, d_Y}} is a continuous bijection, then is f−1f^{-1} also continuous?
Solution.

No. For example, let f ⁣:(R,ddisc)→(R,∣ ⋅ ∣)\func{f}{\p{\R, d_{\mathrm{disc}}}}{\p{\R, \abs{\:\cdot\:}}} be given by f(x)=xf\p{x} = x. Then ff is continuous since every function on a discrete metric space is continuous, but its inverse f−1 ⁣:(R,∣ ⋅ ∣)→(R,ddisc)\func{f^{-1}}{\p{\R, \abs{\:\cdot\:}}}{\p{\R, d_{\mathrm{disc}}}} is not:

Let V={0}V = \set{0}, which is open since every set in a discrete metric space is open. But

(f−1)−1(V)=f({0})={0}\p{f^{-1}}^{-1}\p{V} = f\p{\set{0}} = \set{0}

is not open with respect to the absolute value metric. This means that f−1f^{-1} is not continuous.

However, with a few more assumptions on XX, we can guarantee that f−1f^{-1} is automatically continuous:

Example 5.

Let f ⁣:(X,dX)→(Y,dY)\func{f}{\p{X, d_X}}{\p{Y, d_Y}} be a continuous function. If XX is compact, then f−1f^{-1} is also continuous.
Solution.

We're going to prove this by using (iv). Let F⊆XF \subseteq X be a closed set. Since XX is compact, this means that FF is also compact, so because ff is continuous,

(f−1)−1(F)=f(F)⊆Y\p{f^{-1}}^{-1}\p{F} = f\p{F} \subseteq Y

is also compact, hence closed. In other words, the preimage of every closed set in XX under f−1f^{-1} is closed, so f−1f^{-1} is continuous.