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In a general topological space $(X,\tau)$ I have the following situation: $$F\subset M\subset N$$.

If I prove that $F$ is compact in $N$ (w.r.t the induced topology), is it true that $F$ is compact in $M$ too (w.r.t the induced topology)?

Can I apply this result to the following exercise?

Let $F\subset (C^1([a,b]),\lVert\,\cdot\,\rVert_{\infty})$. If I prove that $F$ is compact in $(C^1([a,b]),\lVert\,\cdot\,\rVert_{C^1})$, is it true that $F$ is automatically compact in $(C^1([a,b]),\lVert\,\cdot\,\rVert_{\infty})$?

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  • $\begingroup$ So, do you know whether $(C^1([a,b]),\lVert\,\cdot\,\rVert_{\infty})$ is a subset and a topological subspace of $(C^1([a,b]),\lVert\,\cdot\,\rVert_{C^1})$? $\endgroup$ – MPW May 7 '14 at 10:40
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It is meaningless to say that $F$ is compact "in" $N$. Compactness is a property of the space $F$, not of any space in which it happens to be embedded.

As long as the subspace topology induced on $F$ is the same in both cases, it doesn't matter. $F$ is either compact or not, period.

For your example, it will depend on whether or not the inherited topology is the same in both cases. I don't know that offhand.

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Images of compact sets under continuous maps are still compact, whether or not the map is bijective or a homeomorphism to its image or any such details. Continuous images of compacts are compact.

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I think what you are missing is the fact that "induced topology" is transitive. For example, if $A \subset B \subset X$ then the topology on $A$ that is induced from the topology on $B$ that is induced from the topology on $X$ is equal to the topology on $A$ that is induced from the topology on $X$.

It follows that in your situation $F \subset M \subset N \subset X$, as long as all of the topologies on $F,M,N$ are induced from $X$ then all the topologies on $F$ induced from any of $M,N,X$ are equal. And therefore any topology property, such as compactness, is the same for $F$ regardless of which of $M,N,X$ you choose to induce from.

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  • $\begingroup$ So, how can I answer to "Let $F\subset (C^1([a,b]),\lVert\,\cdot\,\rVert_{\infty})$. If I prove that $F$ is compact in $(C^1([a,b]),\lVert\,\cdot\,\rVert_{C^1})$, is it true that $F$ is automatically compact in $(C^1([a,b]),\lVert\,\cdot\,\rVert_{\infty})$?" $\endgroup$ – avati91 May 7 '14 at 12:45

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