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In their paper "The classical Plateau problem and the topology of 3-manifolds", Meeks and Yau claim that for any fixed closed Riemannian 3-manifold $M$ there is a lower bound for the area of non-trivial two-spheres. Furthermore, the least area such is embedded.

A sketch of the desired proof:

We first induct on $s(M)$: the number of essential spheres in a maximal system in $M$. If $s(M) = 0$, then $M$ has universal cover $R^3$ or $S^3$, and we are done, using the proof of Alexander's theorem, say. (This step is not so obvious, but let's move along.)

Suppose that $S$ is a minimal area essential sphere in $M$, and so is embedded. Let $T$ be the given sphere with small area, which is necessarily inessential. If $T \cap S$ is not generic, then move $T$ slightly to make it so. We now induct on $|T \cap S|$. If $T \cap S = \emptyset$ then we can cut along $S$, cap off with a ball, and reduce $s(M)$.

If not, then let $\alpha$ be an innermost curve of intersection. Thus $\alpha$ bounds a disk in $S$ that has area less than either disk it bounds in $T$. (This uses the fact that a surgery of an essential sphere yields at least one essential sphere, and the minimality of $S$.) So we may surger $T$ to get a pair of inessential spheres. (This is because both of them have area less than that of $T$.) This reduces $|T \cap S|$ and we are done.

In their paper "The classical Plateau problem and the topology of 3-manifolds", Meeks and Yau claim that for any fixed closed Riemannian 3-manifold $M$ there is a lower bound for the area of non-trivial two-spheres. Furthermore, the least area such is embedded (or double covers an essential $RP^2$.)

There are many ways to use this to answer the question. Here is one possibility:

We first induct on $s(M)$: the number of essential spheres in a maximal system in $M$. If $s(M) = 0$, then $M$ has universal cover $R^3$ or $S^3$, and we are done, using the proof of Alexander's theorem, say. (This step is not obvious, but let's move along.)

Suppose that $S$ is a minimal area essential sphere in $M$, and so is embedded. Let $T$ be the given sphere with small area, which is necessarily inessential. If $T \cap S$ is not generic, then move $T$ slightly to make it so. We now induct on $|T \cap S|$. If $T \cap S = \emptyset$ then we can cut along $S$ and cap off with a ball to get a manifold $M'$. Note that $s(M') < s(M)$. (We must also check that the area of a smallest essential sphere in $M'$ is greater than the area of $S$.)

If $|T \cap S| \neq \emptyset$, then let $\alpha$ be an innermost curve of intersection. Thus $\alpha$ bounds a disk in $S$ that has area less than either disk it bounds in $T$. (This uses the fact that a surgery of an essential sphere yields at least one essential sphere, and the minimality of $S$.) So we may surger $T$ to get a pair of inessential spheres. (This is because both of them have area less than that of $T$.) This reduces $|T \cap S|$ and we are done.

In their paper "The classical Plateau problem and the topology of 3-manifolds", Meeks and Yau claim that for any fixed closed Riemannian 3-manifold $M$ there is a lower bound for the area of non-trivial two-spheres. Furthermore, the least area such is embedded.

A sketch of the desired proof:

We first induct on $s(M)$: the number of essential spheres in a maximal system in $M$. If $s(M) = 0$, then $M$ has universal cover $R^3$ or $S^3$, and we are done using the proof of Alexander's theorem, say.

Suppose that $S$ is a minimal area essential sphere in $M$, and so is embedded. Let $T$ be the given sphere with small area, which is necessarily inessential.
If $T \cap S$ is not generic, then move $T$ slightly to make it so. We now induct on $|T \cap S|$. If $T \cap S = \emptyset$ then we can cut along $S$, cap off with a ball, and reduce $s(M)$.

If not, then let $\alpha$ be an innermost curve of intersection. Thus $\alpha$ bounds a disk in $S$ that has area less than either disk it bounds in $T$. (This uses the fact that a surgery of an essential sphere yields at least one essential sphere, and the minimality of $S$.) So we may surger $T$ to get a pair of inessential spheres. (This is because both of them have area less than that of $T$.) This reduces $|T \cap S|$ and we are done.

In their paper "The classical Plateau problem and the topology of 3-manifolds", Meeks and Yau claim that for any fixed closed Riemannian 3-manifold $M$ there is a lower bound for the area of non-trivial two-spheres. Furthermore, the least area such is embedded.

A sketch of the desired proof:

We first induct on $s(M)$: the number of essential spheres in a maximal system in $M$. If $s(M) = 0$, then $M$ has universal cover $R^3$ or $S^3$, and we are done, using the proof of Alexander's theorem, say. (This step is not so obvious, but let's move along.)

Suppose that $S$ is a minimal area essential sphere in $M$, and so is embedded. Let $T$ be the given sphere with small area, which is necessarily inessential. If $T \cap S$ is not generic, then move $T$ slightly to make it so. We now induct on $|T \cap S|$. If $T \cap S = \emptyset$ then we can cut along $S$, cap off with a ball, and reduce $s(M)$.

If not, then let $\alpha$ be an innermost curve of intersection. Thus $\alpha$ bounds a disk in $S$ that has area less than either disk it bounds in $T$. (This uses the fact that a surgery of an essential sphere yields at least one essential sphere, and the minimality of $S$.) So we may surger $T$ to get a pair of inessential spheres. (This is because both of them have area less than that of $T$.) This reduces $|T \cap S|$ and we are done.

In their paper "The classical Plateau problem and the topology of 3-manifolds", Meeks and Yau claim that for any fixed closed Riemannian 3-manifold $M$ there is a lower bound for the area of non-trivial two-spheres. Furthermore, the least area such is embedded.

In their paper "The classical Plateau problem and the topology of 3-manifolds", Meeks and Yau claim that for any fixed closed Riemannian 3-manifold $M$ there is a lower bound for the area of non-trivial two-spheres. Furthermore, the least area such is embedded.

A sketch of the desired proof:

We first induct on $s(M)$: the number of essential spheres in a maximal system in $M$. If $s(M) = 0$, then $M$ has universal cover $R^3$ or $S^3$, and we are done using the proof of Alexander's theorem, say.

Suppose that $S$ is a minimal area essential sphere in $M$, and so is embedded. Let $T$ be the given sphere with small area, which is necessarily inessential.
If $T \cap S$ is not generic, then move $T$ slightly to make it so. We now induct on $|T \cap S|$. If $T \cap S = \emptyset$ then we can cut along $S$, cap off with a ball, and reduce $s(M)$.

If not, then let $\alpha$ be an innermost curve of intersection. Thus $\alpha$ bounds a disk in $S$ that has area less than either disk it bounds in $T$. (This uses the fact that a surgery of an essential sphere yields at least one essential sphere, and the minimality of $S$.) So we may surger $T$ to get a pair of inessential spheres. (This is because both of them have area less than that of $T$.) This reduces $|T \cap S|$ and we are done.