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Let $X \subseteq \mathbb R$ and $X$ has same cardinality as $\mathbb R$ , does there always exist a continuous surjection from $\mathbb R$ onto $X$ ? ( I know that there need not always be a continuous surjection from $X$ onto $\mathbb R$ , for example when $X$ is closed bounded interval )

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No. For instance, take $X = \mathbb R \setminus \mathbb Q$, the irrationals. Then $X$ is uncountable and totally disconnected, i.e. its connected components are single points. Since $\mathbb R$ is connected, any continuous map $f: \mathbb R \to X$ sends $\mathbb R$ into one connected component of $X$. Thus $f(\mathbb R)$ is a single point, hence $f$ cannot be surjective.

In fact, this works for any disconnected subset of $\mathbb R$ which is uncountable. Just take $X = \mathbb R \setminus \{0\}$, for instance.

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Not even close. There are $2^{\aleph_0}$ continuous functions from $\Bbb R$ to itself, so there are only $2^{\aleph_0}$ continuous images of $\Bbb R$.

However there are $2^{2^{\aleph_0}}$ subsets of $\Bbb R$ which have the same cardinality as $\Bbb R$ itself. So by cardinality argument alone we see that most sets cannot be obtained in such way.

Indeed, the easy argument as given by Alex G. in the other answer is that a continuous image of $\Bbb R$ must be connected, so any disconnected subset cannot be obtained like that. But what if we replace $\Bbb R$ by a Borel set instead? There are totally disconnected Borel sets, like the irrationals or Cantor sets.

Then the cardinality argument shows that even in that case, most sets are not obtained by a continuous image of a Borel set.

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