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Proving that the closure of a connected set in a metric space is connected.

This was a proof that was done in class and I'm sure that it is wrong

Proof: Let $(X, d)$ be a metric space. Suppose that $Y \subseteq X$ is connected. We show that $\overline{Y}$ is also connected.

Suppose that $(\overline{Y}, d)$ was disconnected, then there exist open sets $U, V$ of $\overline{Y}$ such that $U \cup V = \overline{Y}$ and $U \cap V = \emptyset$ and $U \neq \emptyset$ and $V \neq \emptyset$.

Since $Y \subseteq \bar{Y}$ we thus have $U \cap Y$ and $V \cap Y$ to be clopen in $Y$ and $(U \cap Y) \cup (V \cap Y) = Y$. Assume that $U \cap Y \neq \emptyset$ then if $V \cap Y \neq \emptyset$ we have a contradiction.

So if $U \cap Y \neq \emptyset$ then we must have $V \cap Y = \emptyset$ and $U \cap Y = Y$. Since $V \cap Y =\emptyset$ we must have that $V$ contains only limit points of $Y$.

But then for any $x \in V$ we have $B(x, r) \cap U \neq \emptyset$ for any $r > 0$. So any ball $B(x, r)$ contains points outside of $V$ contradicting the fact that $V$ is open in $(\bar{Y}, d)$. $\square$


I'm a bit confused by the last two lines. $V$ being open in $\bar{Y}$ means $V = \bar{Y} \cap A$ for some open $A$ in $(X, d)$, I don't see why any open ball $B(x, r)$ containing points outside of $V$ would contradict the fact that $V$ is open in $(\bar{Y}, d)$ (It sure would contradict the fact that $V$ is open in $(X, d)$ though)

For example take $(\mathbb{R}, d)$ with the usual metric, and consider $\bar{Y} = [0, 1]$, and the open set $V = [0, \frac{1}{2})$ in $\bar{Y}$ then $B(0, r) = (-r, r)$ contains points outside of $V$ so by the above logic $V$ cannot be open in $\bar{Y}$ (which is wrong because $V$ is certainly open in $\bar{Y}$)

Am I correct in saying that there is an error in this proof?

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  • $\begingroup$ I think the intention is that $B(x,r)$ contains points in $\bar{Y} \setminus V$ (since they are in $U$), giving the contradiction. $\endgroup$ – B. Mehta May 14 '18 at 19:34
  • $\begingroup$ The ball $B(x,r)$ is taken in $\overline{Y}$, not in $X$. This is not explicit but that's the only way. You can even forget about $Y$ and $X$ at all and restate the thesis as: if $X$ has a dense connected subset then it is connected. $\endgroup$ – freakish May 14 '18 at 19:39
  • $\begingroup$ @freakish When you say the ball $B(x, r)$ is taken in $\bar{Y}$" and not in $X$ do you mean that we should take it as $B(x, r) \cap \bar{Y} \neq \emptyset$? Then any neighbourhood (in $\bar{Y}$) of $x \in V$ contains points of $U$ which are outside $V$ giving us the contradiction? $\endgroup$ – Perturbative May 14 '18 at 20:02
  • $\begingroup$ @Perturbative Yes, exactly. $\endgroup$ – freakish May 14 '18 at 20:21
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The proof you quote makes it needlessly complicated: we know $U \cap Y = Y$, $Y \cap V = \emptyset$.

continuing: $V$ being open in $\overline{Y}$ means that there exists $V'$ open in $X$ with $V' \cap \overline{Y} = V$.

Note that $Y \cap V =\emptyset$ implies $Y \subseteq X\setminus V'$ (otherwise for some $y$, $y \in Y \cap V' \subseteq \overline{Y} \cap V' = V$, which is not the case), and so $\overline{Y} \subseteq X\setminus V'$, (as the latter set is closed) and so $V' \cap \overline{Y} = V = \emptyset$. Contradiction and done.

No need to mention balls at all; this works in any topological space.

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The proof is valid but writing $B(x, r) \cap U \neq \emptyset$ is a bit confusing. To clarify the argument, I would replace the last paragraph of the proof with this:

But then for any $x \in V$ we have $B(x, r) \cap Y \neq \emptyset$ for any $r > 0$. So any ball $B(x, r)$ contains points outside of $V$ since $Y \subset U$ and $U \cap V = \emptyset$. But this is absurd since $V$ is open in $(\bar{Y}, d)$.

We also get a contradiction if we assume that $V \cap Y \neq \emptyset$, so it is not possible to disconnect $\bar{Y}$ with two open sets $U$ and $V$. $\square$

Note: We make use of the fact that any point $w$ in an open set $W$ of a metric space belongs to an open ball $B(w, r)$ that is wholly contained in $W$.


So to show $\bar{Y}$ is connected the above examines opens sets. We can argue by focusing on just the closure operator used in topology.

In what follows $X$ is assumed to be a disconnected space with closed sets $K$ and $L$ providing a non-trivial decomposition.

Now any subspace of $T$ of $X$ with points in both $K$ and $L$ is manifestly disconnected along with its closure $\bar{T}$.

If $S$ is a subspace of $X$ that is contained in, say $K$, then $\bar{S} \subset K$.

Suppose we can write $X = \bar{Y}$. Then $Y$ can't be connected. If it was connected it would have to be wholly contained in either $K$ or $L$. In one case we would have that $X = K$ and in other case $X = L\,$. But either one of these identities contradicts the assumption that we started with a non-trivial decomposition of $X$.

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