Uniformly continuous mapping
Posted: 05/10-2011 20:37
Hei.
Setter stor pris på kommentarer/innspill til en oppgave jeg har forsøkt å løse i dag. Oppgaven lyder:
Suppose [tex]f[/tex] is a uniformly continuous mapping of a metric space [tex]X[/tex] into a metric space [tex]Y[/tex] and prove that [tex]\{f(x_n)\}[/tex] is a Cauchy sequence in [tex]Y[/tex] for every Cauchy sequence [tex]\{x_n\}[/tex] in [tex]X[/tex].
LØSNING:
Suppose [tex]f[/tex] is uniformly continuous. We then have that for every [tex]\epsilon > 0[/tex] there exist a [tex]\delta > 0[/tex] such that [tex]d_y(f(p), f(y)) < \epsilon[/tex] for all [tex]p[/tex] and [tex]q[/tex] in [tex]X[/tex] for which [tex]d_x(p, q) < \delta[/tex].
Now suppose we pick an arbitary Cauchy sequence [tex]\{x_n\}[/tex] in [tex]X[/tex]. We then have that for every [tex]\gamma > 0[/tex], there exist an integer [tex]N[/tex] such that [tex]d(p_n, p_m) < \gamma[/tex] if [tex]n \geq N[/tex] and [tex]m \geq N[/tex]. Thus, by setting [tex]\gamma = \delta[/tex], we have that [tex]d_x(p_n, p_m) < \delta[/tex]. Due to uniform continuity, it follows that [tex]d_y(f(p_n), f(p_m)) < \epsilon[/tex]. Since the Cauchy sequence was chosen arbitrarily, it follows that [tex]\{f(x_n)\}[/tex] is a Cauchy sequence in [tex]Y[/tex] for every Cauchy sequence [tex]\{x_n\}[/tex] in [tex]X[/tex] as long as we set [tex]\gamma = \delta[/tex].
Som sagt, kommentarer/innspill er svært velkommen!
Setter stor pris på kommentarer/innspill til en oppgave jeg har forsøkt å løse i dag. Oppgaven lyder:
Suppose [tex]f[/tex] is a uniformly continuous mapping of a metric space [tex]X[/tex] into a metric space [tex]Y[/tex] and prove that [tex]\{f(x_n)\}[/tex] is a Cauchy sequence in [tex]Y[/tex] for every Cauchy sequence [tex]\{x_n\}[/tex] in [tex]X[/tex].
LØSNING:
Suppose [tex]f[/tex] is uniformly continuous. We then have that for every [tex]\epsilon > 0[/tex] there exist a [tex]\delta > 0[/tex] such that [tex]d_y(f(p), f(y)) < \epsilon[/tex] for all [tex]p[/tex] and [tex]q[/tex] in [tex]X[/tex] for which [tex]d_x(p, q) < \delta[/tex].
Now suppose we pick an arbitary Cauchy sequence [tex]\{x_n\}[/tex] in [tex]X[/tex]. We then have that for every [tex]\gamma > 0[/tex], there exist an integer [tex]N[/tex] such that [tex]d(p_n, p_m) < \gamma[/tex] if [tex]n \geq N[/tex] and [tex]m \geq N[/tex]. Thus, by setting [tex]\gamma = \delta[/tex], we have that [tex]d_x(p_n, p_m) < \delta[/tex]. Due to uniform continuity, it follows that [tex]d_y(f(p_n), f(p_m)) < \epsilon[/tex]. Since the Cauchy sequence was chosen arbitrarily, it follows that [tex]\{f(x_n)\}[/tex] is a Cauchy sequence in [tex]Y[/tex] for every Cauchy sequence [tex]\{x_n\}[/tex] in [tex]X[/tex] as long as we set [tex]\gamma = \delta[/tex].
Som sagt, kommentarer/innspill er svært velkommen!