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iHav to Drive
Eastern Imports
Let X := {(a(n)): summation|a(n)| < infinity{ be the space of absolutely convergent
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<blockquote data-quote="LeoL" data-source="post: 1437821" data-attributes="member: 246955"><p><strong>Let X := {(a<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down (n)" data-smilie="23"data-shortname="(n)" />): summation|a<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down (n)" data-smilie="23"data-shortname="(n)" />| < infinity{ be the space of absolutely convergent</strong></p><p></p><p>sequences. Define the? l1 and l(infinity) metrics on this space by</p><p>d(l1)(a<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down (n)" data-smilie="23"data-shortname="(n)" />, b<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down (n)" data-smilie="23"data-shortname="(n)" />) := summation |a<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down (n)" data-smilie="23"data-shortname="(n)" /> - b<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down (n)" data-smilie="23"data-shortname="(n)" />|</p><p>where n=0 at first and goes to n = infinity.</p><p>d(l(infinity))(a<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down (n)" data-smilie="23"data-shortname="(n)" />, b<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down (n)" data-smilie="23"data-shortname="(n)" />) := sup|a<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down (n)" data-smilie="23"data-shortname="(n)" /> - b<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f44e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt="(n)" title="Thumbs down (n)" data-smilie="23"data-shortname="(n)" />|</p><p></p><p>I hope this is clear if not let me know. Show that there exist sequences x(of elements of X which are convergent with respect to the d(l(infinity)) metric but not with respect to the d(l1) metric</p></blockquote><p></p>
[QUOTE="LeoL, post: 1437821, member: 246955"] [b]Let X := {(a(n)): summation|a(n)| < infinity{ be the space of absolutely convergent[/b] sequences. Define the? l1 and l(infinity) metrics on this space by d(l1)(a(n), b(n)) := summation |a(n) - b(n)| where n=0 at first and goes to n = infinity. d(l(infinity))(a(n), b(n)) := sup|a(n) - b(n)| I hope this is clear if not let me know. Show that there exist sequences x(of elements of X which are convergent with respect to the d(l(infinity)) metric but not with respect to the d(l1) metric [/QUOTE]
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