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Convergence in the uniform metric is exactly uniform convergence: one $N$ serving every point

Statement

Let $X$ be a nonempty set, let $(Y, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), and let $\overset{ˉ}{ρ}$ be the uniform metric on $Y^{X}$ (For a nonempty set $X$ and a metric space $(Y, d)$ the uniform metric $\overset{ˉ}{ρ} (f, g) = \sup_{x} \min {d (f (x), g (x)), 1}$ is a metric on $Y^{X}$ ). Let $(f_{k})$ be a sequence in $Y^{X}$ and let $f \in Y^{X}$ . Then

$f_{k} \to f in (Y^{X}, \overset{ˉ}{ρ}) ⟺ (f_{k}) converges uniformly to f,$

convergence in a metric space being Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ and uniform convergence being Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^{X}$ and on $C (X, Y)$ .

This is what makes the name of the topology accurate, and it is the reason the truncation at $1$ in the uniform metric costs nothing: below the threshold the truncated and untruncated distances agree, and convergence is a statement about arbitrarily small distances. No choice principle is used.

Facts & Assumptions

Given: A nonempty set $X$ , a metric space $(Y, d)$ , the truncated metric $\overset{ˉ}{d} = \min {d, 1}$ on $Y$ , the uniform metric $\overset{ˉ}{ρ} (g, h) = \sup_{x} \overset{ˉ}{d} (g (x), h (x))$ on $Y^{X}$ , a sequence $(f_{k})$ in $Y^{X}$ and a point $f \in Y^{X}$ .

[L1]

$\overset{ˉ}{d} (u, v) \leq d (u, v)$ and $\overset{ˉ}{d} (u, v) \leq 1$ for all $u, v \in Y$ , the minimum of a two-element set of reals being a lower bound of both elements and one of them ( $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L2]

If $\overset{ˉ}{d} (u, v) < 1$ then $\overset{ˉ}{d} (u, v) = d (u, v)$ : the minimum $\min {d (u, v), 1}$ is one of its two arguments, and it is not $1$ , so it is $d (u, v)$ (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology).

[L3]

$\overset{ˉ}{ρ} (g, h)$ is an upper bound of ${\overset{ˉ}{d} (g (x), h (x)) : x \in X}$ and is the least one; in particular $\overset{ˉ}{d} (g (x), h (x)) \leq \overset{ˉ}{ρ} (g, h)$ for every $x \in X$ , and any real bounding all these values above bounds $\overset{ˉ}{ρ} (g, h)$ (For a nonempty set $X$ and a metric space $(Y, d)$ the uniform metric $\overset{ˉ}{ρ} (f, g) = \sup_{x} \min {d (f (x), g (x)), 1}$ is a metric on $Y^{X}$ , Complete ordered field (least-upper-bound property), Suprema and infima are unique).

[L4]

$g_{k} \to g$ in a metric space means: for every rational $ε > 0$ there is $K \in N$ with the distance from $g_{k}$ to $g$ below $ε$ for every $k \geq K$ ; and the test with a real $ε > 0$ is equivalent, since below every positive real lies a positive rational (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , The rationals embed densely in the reals, Open ball, closed ball and sphere in a metric space).

[L5]

The minimum of two positive reals is positive, and halving a positive real gives a positive real strictly below it (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1assume-hyp

Suppose $(f_{k})$ converges uniformly to $f$ , and let $ε > 0$ be real.

1.2assume-hyp

Suppose instead that $f_{k} \to f$ in $(Y^{X}, \overset{ˉ}{ρ})$ , and let $ε > 0$ be real.

2.1step 1.1L5choose

Under step 1.1: put $η : = ε / 2$ , a real with $0 < η < ε$ , and take $K \in N$ with $d (f_{k} (x), f (x)) < η$ for every $x \in X$ and every $k \geq K$ .

2.2step 1.2L4L5choose

Under step 1.2: put $η : = \min {ε, 1} / 2$ , a real with $0 < η \leq 1 / 2 < 1$ and $η < ε$ , and take $K \in N$ with $\overset{ˉ}{ρ} (f_{k}, f) < η$ for every $k \geq K$ .

3.1step 2.1L1L3

Under step 1.1: for $k \geq K$ and every $x \in X$ we have $\overset{ˉ}{d} (f_{k} (x), f (x)) \leq d (f_{k} (x), f (x)) < η$ , so $η$ bounds that set of values above and hence $\overset{ˉ}{ρ} (f_{k}, f) \leq η < ε$ .

3.2step 2.2L2L3

Under step 1.2: for $k \geq K$ and every $x \in X$ we have $\overset{ˉ}{d} (f_{k} (x), f (x)) \leq \overset{ˉ}{ρ} (f_{k}, f) < η < 1$ , so $\overset{ˉ}{d} (f_{k} (x), f (x)) = d (f_{k} (x), f (x))$ and therefore $d (f_{k} (x), f (x)) < η < ε$ .

4.1step 3.1L4

Step 3.1 produces, for each real $ε > 0$ , an index $K$ with $\overset{ˉ}{ρ} (f_{k}, f) < ε$ for every $k \geq K$ , which is convergence $f_{k} \to f$ in $(Y^{X}, \overset{ˉ}{ρ})$ ; this is the forward implication.

4.2step 3.2

Step 3.2 produces, for each real $ε > 0$ , an index $K$ with $d (f_{k} (x), f (x)) < ε$ for every $x \in X$ and every $k \geq K$ , which is uniform convergence of $(f_{k})$ to $f$ ; this is the converse implication.

5.1step 4.1step 4.2∎

Steps 4.1 and 4.2 are the two implications, so the two conditions are equivalent.

Remarks

Where the threshold $1$ enters and where it does not. It enters only in step 3.2, which needs the distance to be strictly below $1$ before the truncation can be undone; that is arranged by shrinking $η$ to at most $1 / 2$ , which costs nothing because $η$ is being made small anyway. It does not enter the forward direction at all, since $\overset{ˉ}{d} \leq d$ outright.
The lemma fails for the value of the distance, not for convergence. The numbers $\overset{ˉ}{ρ} (f, g)$ and $\sup_{x} d (f (x), g (x))$ differ as soon as some distance exceeds $1$ , and the second need not exist. What the lemma says is that the two determine the same convergent sequences and the same limits, which is all a topology sees.
Uniform convergence implies pointwise convergence, and not conversely. From the definition, an index serving every point serves each point separately, so a uniformly convergent sequence converges at every point (A sequence converges in the topology of pointwise convergence exactly when it converges at every point). The converse fails, and the companion page exhibits the standard witness on $[0, 1]$ .

Convergence in the uniform metric is exactly uniform convergence: one $N$ serving every point

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Convergence in the uniform metric is exactly uniform convergence: one NN serving every point

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Convergence in the uniform metric is exactly uniform convergence: one $N$ serving every point