https://topospaces.subwiki.org/w/index.php?title=Local_compactness_is_weakly_hereditary&feed=atom&action=history Local compactness is weakly hereditary - Revision history 2022-06-27T08:06:35Z Revision history for this page on the wiki MediaWiki 1.29.2 https://topospaces.subwiki.org/w/index.php?title=Local_compactness_is_weakly_hereditary&diff=2636&oldid=prev Vipul: New page: {{topospace metaproperty satisfaction}} ==Statement== ===Property-theoretic statement=== The property of topological spaces of being locally compact satisf... 2008-07-21T20:35:37Z <p>New page: {{topospace metaproperty satisfaction}} ==Statement== ===Property-theoretic statement=== The <a href="/wiki/Property_of_topological_spaces" title="Property of topological spaces">property of topological spaces</a> of being <a href="/wiki/Locally_compact_space" title="Locally compact space">locally compact</a> satisf...</p> <p><b>New page</b></p><div>{{topospace metaproperty satisfaction}}<br /> <br /> ==Statement==<br /> <br /> ===Property-theoretic statement===<br /> <br /> The [[property of topological spaces]] of being [[locally compact space|locally compact]] satisfies the [[metaproperty of topological spaces]] of being [[weakly hereditary property of topological spaces|weakly hereditary]].<br /> <br /> ===Verbal statement===<br /> <br /> Any [[closed subset]] of a [[locally compact space]] is locally compact.<br /> <br /> ==Facts used==<br /> <br /> * [[Compactness is weakly hereditary]]: Any closed subspace of a compact space is compact.<br /> <br /> ==Proof==<br /> <br /> '''Given''': A locally compact space &lt;math&gt;X&lt;/math&gt;, a closed subset &lt;math&gt;C&lt;/math&gt;<br /> <br /> '''To prove''': &lt;math&gt;C&lt;/math&gt; is locally compact<br /> <br /> '''Proof''': We need to show that given any point &lt;math&gt;x \in C&lt;/math&gt;, there exists an open subset containing &lt;math&gt;x&lt;/math&gt; contained in a closed compact subset of &lt;math&gt;C&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;X&lt;/math&gt; is locally compact, there exists an open set &lt;math&gt;V \ni x&lt;/math&gt; and a closed compact subset &lt;math&gt;K&lt;/math&gt; of &lt;math&gt;X&lt;/math&gt; containing &lt;math&gt;V&lt;/math&gt;.<br /> <br /> By the definition of subspace topology, &lt;math&gt;V \cap C&lt;/math&gt; is an open subset of &lt;math&gt;C&lt;/math&gt;. Call this &lt;math&gt;U&lt;/math&gt;. Further, &lt;math&gt;K \cap C&lt;/math&gt; is a closed subset of &lt;math&gt;C&lt;/math&gt;. Call this &lt;math&gt;L&lt;/math&gt;. We then have &lt;math&gt;x \in U \subset L&lt;/math&gt;, with &lt;math&gt;U&lt;/math&gt; open and &lt;math&gt;L&lt;/math&gt; closed.<br /> <br /> We need to show that &lt;math&gt;L&lt;/math&gt; is compact. For this, observe that &lt;math&gt;L = K \cap C&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; is closed in &lt;math&gt;X&lt;/math&gt;, so &lt;math&gt;L \subset K&lt;/math&gt; is closed as a subset of &lt;math&gt;K&lt;/math&gt;. Since any closed subset of a compact space is compact, we conclude that &lt;math&gt;L&lt;/math&gt; is compact, completing the proof.</div> Vipul