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In a paper I am reading, the following is considered obvious:

Let $K$ be a compact and simply connected subset of $\,\mathbb R^2$ and $U\subset \mathbb R^2$ open with $K\subset U$. Then there exists a simply connected and open $V\subset \mathbb R^2$, with $K\subset V\subset U$. More generally, if $K$ is compact, $\mathbb R^2\smallsetminus K$ is connected and $U\subset \mathbb R^2$ open with $K\subset U$, then there exists an open $V\subset \mathbb R^2$, with $K\subset V\subset U$, such that all the connected components of $V$ are simply connected.

I have not managed to see why this is obvious. So far, I have shown this for simply connected compact sets $K$ with sufficiently smooth boundaries.

Any ideas?

EDIT. In all cases, it is assumed that $K$ has connected complement in $\mathbb C$.

In a paper I am reading, the following is considered obvious:

Let $K$ be a compact and connected subset of $\,\mathbb R^2$, with $\mathbb R^2\smallsetminus K$ also connected, and $U\subset \mathbb R^2$ open with $K\subset U$. Then there exists a simply connected and open $V\subset \mathbb R^2$, with $K\subset V\subset U$. More generally, if $K$ is compact, $\mathbb R^2\smallsetminus K$ is connected $($and $K$ not necessarily connected$)$ and $U\subset \mathbb R^2$ open with $K\subset U$, then there exists an open $V\subset \mathbb R^2$, with $K\subset V\subset U$, such that all the connected components of $V$ are simply connected.

I have not managed to see why this is obvious. So far, I have shown this for simply connected compact sets $K$ with sufficiently smooth boundaries.

Any ideas?

In a paper I am reading, the following is considered obvious:

Let $K$ be a compact and simply connected subset of $\,\mathbb R^2$ and $U\subset \mathbb R^2$ open with $K\subset U$. Then there exists a simply connected and open $V\subset \mathbb R^2$, with $K\subset V\subset U$. More generally, if $K$ is compact, $\mathbb R^2\smallsetminus K$ is connected and $U\subset \mathbb R^2$ open with $K\subset U$, then there exists an open $V\subset \mathbb R^2$, with $K\subset V\subset U$, such that all the connected components of $V$ are simply connected.

I have not managed to see why this is obvious. So far, I have shown this for simply connected compact sets $K$ with sufficiently smooth boundaries.

Any ideas?

EDIT. In all cases, it is further assumed that $K$ has connected complement.

In a paper I am reading, the following is considered obvious:

Let $K$ be a compact and simply connected subset of $\,\mathbb R^2$ and $U\subset \mathbb R^2$ open with $K\subset U$. Then there exists a simply connected and open $V\subset \mathbb R^2$, with $K\subset V\subset U$. More generally, if $K$ is compact, $\mathbb R^2\smallsetminus K$ is connected and $U\subset \mathbb R^2$ open with $K\subset U$, then there exists an open $V\subset \mathbb R^2$, with $K\subset V\subset U$, such that all the connected components of $V$ are simply connected.

I have not managed to see why this is obvious. So far, I have shown this for simply connected compact sets $K$ with sufficiently smooth boundaries.

Any ideas?

EDIT. In all cases, it is assumed that $K$ has connected complement in $\mathbb C$.

In a paper I am reading, the following is considered obvious:

Let $K$ be a compact and simply connected subset of $\,\mathbb R^2$ and $U\subset \mathbb R^2$ open with $K\subset U$. Then there exists a simply connected and open $V\subset \mathbb R^2$, with $K\subset V\subset U$. More generally, if $K$ is compact, $\mathbb R^2\smallsetminus K$ is connected and $U\subset \mathbb R^2$ open with $K\subset U$, then there exists an open $V\subset \mathbb R^2$, with $K\subset V\subset U$, such that all the connected components of $V$ are simply connected.

I have not managed to see why this is obvious. So far, I have shown this for simply connected compact sets $K$ with sufficiently smooth boundaries.

Any ideas?

In a paper I am reading, the following is considered obvious:

Let $K$ be a compact and simply connected subset of $\,\mathbb R^2$ and $U\subset \mathbb R^2$ open with $K\subset U$. Then there exists a simply connected and open $V\subset \mathbb R^2$, with $K\subset V\subset U$. More generally, if $K$ is compact, $\mathbb R^2\smallsetminus K$ is connected and $U\subset \mathbb R^2$ open with $K\subset U$, then there exists an open $V\subset \mathbb R^2$, with $K\subset V\subset U$, such that all the connected components of $V$ are simply connected.

I have not managed to see why this is obvious. So far, I have shown this for simply connected compact sets $K$ with sufficiently smooth boundaries.

Any ideas?

EDIT. In all cases, it is further assumed that $K$ has connected complement.